MMW M« TuiOTrriiinTiT jr-iautimmaiitMau COURSE IN Pharmaceutical and Chemical Arithmetic J. W. STURMER MMiMMmnQfl Digitized by tine Internet Archive in 2007 witii funding from IVIicrosoft Corporation littp://www.arcliive.org/details/courseinpliarmaceOOsturuoft COURSE IN Pharmaceutical' and Chemical Arithmetic INCLUDING WEIGHTS AND MEASURES BY JULIUS WILLIAM STURMER, Ph.M., Phar.D. PrefcMor of Pharmaceutical Chemistry, and Dean of Science, Philadelphia College of Pharmacy and Science FOURTH P.BTTjnk^ ' ^ WITH ANSWERS 3^. \ 3 3 PUBLISHED BY THE AUTHOR 1922 PRICE $1.75 Copynght, 1917 By Jxjuus WiLUAu Sturmer PREFACE TO THE FOURTH EDITION. The preceding editions having been welcomed so heartily, the writer deems a radical change in the arrangement or in the scope of the COUESE unnecessary, and no such change has been made in this revision. The advent of a new pharmacopoeia has, however, necessitated a large number of minor chang^es. A short presentation of volumetric calculations has been added; but discussions on intricate volumetric processes— even though these determine the procedure in calctilating the resulta — are considered beyond the scope of this volume. The first four chapters appear elementary, from an arith- metical standpoint. And so they are. But the experience of teachers, of members of boards of pharmacy, and of practical, pharmacists, has proven that, notwithstanding the adoption of the higher entrance requirements by the colleges, instruction covering the ground of these chapters is as necessary to-day as it was in former years. As in the preceding editions, so also in this, the writer has avoided the presentation of a multiplicity of rules and mathe- matical formulas. Indeed, with exception of the arbitrary rule for calculating the doses for children, and the rule concerning the Beaume hydrometer scales, no rules are given which the student is not taught to deduce from solutions of problems in the text. In this edition ANSWERS are given for a suflBcient number of problems throughout the COURSE, so that students studying w^ithout the aid of a teacher may be able to note their progress. But at the request of instructors who use the book in their classes, some problems are left unanswered. It should be stated that some of the answers given are only approximately correct, as in some instances fractions were "rounded off" to shorten the calculations. JULIUS WILLIAM STURMER. CONTENTS. Pages. Weights and Measures 7 A. The Metric System 7-13 Metric Prescriptions 13-16 Metric Prototypes 16-17 Problems and Exercises 17-22 B. English Systems Customary in the United States 22 1. Linear Measure « 22-23 2. Apothecaries' Fluid Measure 24-30 Roman Numerals 25-26 Arithmetical Operations 26-30 8. Imperial Measure 30-31 4. Avoirdupois Weight 31-33 5. Apothecaries' Weight 33-37 Relationship of Systems 38 1. Metric and English Linear Measures 38-39 2. Metric and Apothecaries' Fluid Measures 39-41 3. Apothecaries' and Imperial JNIeasures 41-42 4. Metric and Imperial Measures 42-43 5. Approximate Measures and their Values 43-45 6. Metric and Apothecaries' Weights 45-46 7. Metric and Avoirdupois Weights 47-48 8. Avoirdupois (or Imp.) and Apothecaries' Weights.. 48-49 Volumes and Weights 50 Specific Gravity 50-53 Specific Volume 54-55 Reducing Cubic Centimeters to Grammes 55-56 Reducing Fluid Oimces to Grains 56-57 Reducing Other Units Apoth. Measure to Apoth, Weights 57-59 Reducing Grammes to Cubic Centimeters 59-60 Reducing Grains to Fluid Ounces 60-61 Reducing Grains to Minims 61-62 Drachms to Fl, Drachms and Ounces to Fl. Ounces. 62 Reducing and Enlarging Formulas 63-66 Calculating Definite Weight from Parts by Weight. 65-66 Proportions 67-72 Trade Discount 72 Percentage Problems 73- 87 Volume Percentage 83- 84 Weight to Volume Solutions 87- 88 Other Methods of Expressing Strength of Solutions 88- 89 Concentration and Dilution 90- 94 Pharmacopoeial Rules for Dilution of Alcohol 92 6 Contents Alligation 95 Strength of Mixttrre Calculated from Amounts and Strengths of Ingredients 95- 0(1 Specific Gravity of Mixture Calculated from Amounts and Specific Gravities of Ingredients.. 96- 97 To Calculate Proportion in which Ingredients of Known Strength Should be Mixed to Yield Pro- duct of Required Strength 98-103 In Case Ingredients are of Known Specific Gravity and a Product of Specified Specific Gravity is Wanted 103-104 In Case Amounts of Ingredients are to be Calcu- lated, Either the Strength or the Specific Gravity of Product and of each Ingredient being given. .104-109 Chemical Problems 110 Molecules, Atoms, Atomic Weights, Molecular Weights 110-111 Calculation of Molecular Formula 112 Calculation of Molecular Weight 112-114 Calculation of Percentage Composition from Mol- ecular Formula 114-117 Calculations Based on Chemical Reactions 117-122 The Same, Involving Also Percentage Solutions. .122-125 The Same, Involving Also Conversion from Weight to Volume, or Vice Versa 126-127 Calculations Based on Gravimetric Analyses. . .127-128 Analytical Factor (Gravimetric) 128-129 Calculations Based on Volumetric Analyses. ..129-132 Normal and Fractional Normal Solutions 129-130 Adjusting Strength of Volumetric Solutions 130 Volumetric Factors 131-132 Correction Factor 132 Reducing Volumes of Gases to Weights, and Vice Versa. .133 Effect of Variation in Pressure on the Vol. of a Gasl34-136 Effect of Variation in Temp, on the Vol. of a Gas 136-137 The Same, Involving Also Chemical Reactions. .138 Ihermometer Scales 139 Centigrade to Fahrenheit 140-141 Fahrenheit to Centigrade 141-142 Centigrade and Reaumur 142 Fahrenheit and Reaumur 142 Problems and Answers 143 tliscellaneous Beaumd Hydrometer Scales . 144 Calculating Doses for Children 144 Reference Tables 145 Calculating Weights from Volumes and Vice Versa 145 Values which are Proportional, Table of 145 Table of Atomic Weights 146 CHAPTER I. WEIGHTS AND MEASURES A.— The Metric System. The most important feature of the metric system is that the units of each of its branches are in decimal progression. Ten of any unit make one of the next higher, ten of which in turn make one of a still higher value ; etc. The system thus is in conformity with our arithmetical notation ; and de- nominate numbers in metric units can be added, subtracted, multiplied, and divided as readily as abstract numbers. In this lies the chief practical advantage of the metric system over the other systems in vogue. The branches of the metric system are: (i)linear meas- ure, (2) surface measure, (3) cubic measure (dry measure), (4) cubic measure for liquids (volume measure), (5) and weight. The relationship between these branches is a very simple one — a fact constituting the second great advantage of this system of weights and measures. The primary unit of linear measure — the Meter — is the basis of the entire system. Thus, the primary unit of surface measure, called Are, is the square of ten Meters. The primary unit of cubic measure (dry), the Stere, is the cube of one Meter. The primary unit of volume measure, the Liter, is the cube of one-tenth of a Meter. And the primary unit of weight, the Gramme, is the weight of a cubic r^ Meter* of water, at its greatest density, and weighed in vacuo. Their relationship haa been slightly altered by inaccuracy in the welghinga laade when the first metric weight was constructed. See page 12. 8 Pharmaceutical Arithmetic For convenience in measuring, secondary units, some larger, and some smaller than the units already referred to, were established. These secondary units, as has been stated, differ from each other by ten or by some power of ten. The names for the secondary units are formed by joining certain Greek and Latin prefixes — Greek for multiples, Latin for the fractional — ^to the name of the unit itself. The prefixes used are : Latin Greek milli = thousandth (rdVo) centi = hundredth (i^) deci = tenth (iV) Deka = ten (10) Hecto = hundred (100) Kilo = thousand (1000) Myria = teu-thousand(lOOOO). Thus we have millimeter, milliliter, milligramme ; centi' meter, centiliter, centigramme; etc. Rules for capitalising names of units. — The names of the primary units — Meter, [Are, Stere], Liter, Gramme — are usually written with a capital, as are also the names of the secondary units larger than the primary ones; but the names of the units smaller than the primary units are writ- ten with a small initial letter. Examples: D^^ameter, but rfmmeter; Mynagramme, but milligramme. These rules, as will be seen, serve to decrease the similarity in appearance between the names of certain multiples and certain fraction- als. However, the rules are not universally observed, and in many scientific works the names of all units are written with small initial letters. Abbreviations. — The abbreviations for primary units are : Meter = M ; Liter = L ; Gramme = Gm. The abbreviations for the prefixes are: — milli = m; centi = c ; deci = d ; Deka = D ; Hecto = H ; Kilo = K ; Myria = M. The Metric System 9 To construct the abbreviation of any secondary unit, the abbreviation of the prefix is coupled with the abbrevia- tion for the primary unit. Thus Kilometer = Km. ; milli- meter = mm ; millilitei = ml. ; centigramme = cgm. ; etc However, in case of abbreviations of secondary units oi weight the final m is frequently omitted. See table on page 13- The rules given for capitalizing names of units apply likewise to their abbreviations; Dm. being used for Dekameter, dm. for decimeter, Mgm. for Myriag^amme, and mg^. for milligramme. But in the U. S. Pharmacopoeia the term mil, plural mils, is used in place of ml. In the preceding edition of the Pharmacopoeia (U. S. P., VIII) the mil was called the cubic centimeter, abbreviated c.c, this latter term being still in general use in chemical litera- ture. The mil is not capitalized. The mil and the c.c. are used interchangeably in the succeeding pages. I. Linear Measure. The primary unit of length, the Meter, may be defined as the 4 0.0 0 0.0 0 0' of the earth's circumference, measured across the poles. The distance from the equator to the north pole was calculated from surveys made along the meridian which passes through Paris ; and this distance, di- vided by 10,000,000, was chosen as the unit of length.* Note. — It Is deemed desirable that a system of measure be based upon some unalterable object in nature, in order that the correctness of the measures accepted as models may be re- determined, should this become necessary. The metric system is based upon the dimensions of the earth — an object ill adapted lor a "natural standard;" first, because the earth is not abso- lutely unalterable in size, and second, because of the difficulty of obtaining the measurement. A far better "natural standard" is the second's pendulum, upon the length of which the English measure ^s now based. See page 22. •According to recent measurementa the earth's quadrant measures 10,000,880 Meters, showing: that a slight error had been made in the orig^inal measurements, end that the unit now in use is a trifle shorter than the meter required by theory. 10 Pharmaceutical Arithmetic Table of Linear Measure. Table A. 10 millimeters (mm.) = i centimeter (cm.) 10 centimeters = i decimeter (dm.) 10 decimeters = i Meter (M.) lo Meters = i Dekameter (Dm.) lo Dekameters = i Hectometer (Hm.) lo Hectometers = i Kilometer (Km.) 10 Kilometers = i Myriameter (Mm.) For miscroscopic measurements the micro-millimeter, which is xtiVt of a millimeter, is used. It is usually writ- ten micron or mikron, and abbreviated mkm. Table B. Kilometer Hectometers Dekameters Meters decimeters centimeters millimeters 1 = 10 = 100 = 1000 = 10000 = 100000 = 1000000 1 == 10 = 100 = 1000 = 10000 = lOCOOO 1 = 10= 100= 1000= 10000 1= 10= 100= 1000 1 = 10 = 10« 1= 10 Units in Common Use. — Units larger than the Meter are seldom employed in pharmaceutical literature. As has been stated on page 9, the U. S. Pharmacopoeia ignores the larger units, and in its text mm. and cm. are capitalized without danger of ambiguity. 2. Surface Measurjs. 3. Cubic Measure for SoLros. [Dry Measure.] These two branches of the metric system are not used in pharmacy ; hence are here omitted. 4. Cubic Measure for Fluids. [Volume Measure.] For a primary unit of volume the cube of r\> Meter was The Metric System ii selected, and was named Liter [pronounced Leeter, not Laiter.]* But the present Liter is the volume of i Kg. of water, at 4''C., and under standard atmospheric pressure. Tables of Volume Measure. Table A. 10 milliliters (mils) = i centiliter (cl.) ID centiliters = I deciliter (dl.) ID deciliters = I Liter (L.) 10 Liters = I Dekaliter (Dl.) 10 Dekaliters = I Hectoliter (HI.) 10 Hectoliters = I Kiloliter (Kl.) ID Kiloliters = I Myrialiter (Ml). The milliliter (mil) is also called cubic centimeter, to which it bears the same relation that the Liter bears to the cubic decimeter. The abbreviation for cubic centimeter is c.c, c.cm., or, as in the old U. S. P., Cc. Table B. MvrtalUer KtioWers HektolUers DehUUers UUrt deeUUers eentilUers mOUlUers 1 == 10 = 100 = 1000 = 10000 = 100000 = 1000000 = 10000000 1 = 10 = 100 = 1000 = 10000 = 100000 = 1000000 1 = 10 = 100 = 1000 = 10000 = 100000 1 = 10 = 100 = 1000 = 10000 1 = 10 = 100 = 1000 1 -I 10 = 100 1 = 10 Units in Use. — In pharmaceutical and chemical works only two units are commonly used : the Liter, and the milli- liter, which latter, however, is called the mil or the cubic ♦ The more refined methods of weighing and measuring of recent Inves- tigators have brought to light the fact that the Liter In actual use differs slightly from the Liter required by theory. 12 Pharmaceutical Arithmetic centimeter. The table of volume measure, as actually used is therefore very simple; namely looo mils (cubic centimeters) = I Liter. However, in books and periodicals the other units are some- times met with, and therefore should not be overlooked by the student. 5. Weight. The primary unit of weight, the Gramme, is defined as the weight of one mil of water, at 4*'C., and under standard atmospheric pressure.* In pharmacy the name of the primary unit of weight is always spelled according to French orthography — Gramme. The English spelling, gram, gives to the word too great a similarity to the word grain — the name of a unit of weight of the Apothecaries' System of Weight. The pronunciation is gram, the a having the same sound as in Sam. The abbre- viation is always capitalized, Gm., lest it be mistaken for gr., which is the abbreviation for g^ain. *Weig:ht in a vacuum is called tmt weight, in contradistinction to apparmi tceight, which ifl weight obtained in air, i. e., under ordinary conditions. Air, like water, though in a leaser degree, exerts a buoyant force on bodies; tending to lift or float both the body being weighed and the weights used. This buoyant force is exerted on a body in proportion to its bulk. Hence, if there is a difference in bulk between the body weighed and the weights used, there will be a corresponding difference in this buoyant force, and the weight obtained will not be truly proportional to masa It follows also that this difference in buoyant force will fluctuate with the density of the air, i. e., with the barometric pressure. For this reason the ba- rometer reading is taken into consideration in weighings where great accuracy is required, and weighing in yacuo la impracticable. See page 31. Recent determination of the weight of water under standard condi- tions, and witk the most delicate balances, hare shown that the original Eilo- grramne weight, which has serred as the model for orer 100 years, is about .58 Grammes too light to represent the weight of 1 cubic decimeter of water. The primary unit itself, the Gramme, is, therefore, .00058 Gm. or nearly .06 per cent. ilghter than was intended. But as the units of Tolume were made correspondingly smaller one milliliter (actual) being the volume of one Gramme (actual) ot water, and one Liter (actual), the voltune of one Kilogramme (actual), this error In the construction of the original Eilogramme weight haa no practical bearinf;. The Metric System 13 Table A. 10 milligrammes (mg.)= i centigramme (cgm. or eg.) 10 centigrammes = i decigramme (dgm. or dg,) 10 decigrammes = i Gramme (Gm.) 10 Grammes = i Dekagramme (Dgm. or Dg.) 10 Dekagrammes = i Hectogramme (Hgm. or Hg.) 10 Hectogrammes =: I Kilogramme (Kgm. or Kg.) 10 Kilogrammes = i Myriagramme (Mgm.) As already stated, the final m in these abbreviations is frequently omitted. Table B. ^fyria Kilo Hecto Deka Grammes deci centi mini gramme grammes grammes grammes grammes grammes grammes 1 10 = 100 = 1000 = 10000 = 100000 = 1000000 = 10000000 1 = to = 100 = 1000 = 10000 = 100000 = 1000000 = 10 = 100 = 1000 = 10000 = 100000 1 = 10 = 100 = 1000 = 10000 1 = 10 = 1 100 = 10 = 1 = 1000 100 10 Units in Common Use. — In prescriptions, and in the U. S. Pharmacopoeia, only one unit of weight — the Gramme — is used. In many scientific books and periodicals- the deci- gramme, centigramme, and milligramme are used in addi- tion. The Kilogramme is used in commercial transactions, and is frequently abbreviated to Kilo, pronounced Killo, not Kailo. Me^tric Prescriptions. Rules. — I. All volumes should be expressed in cubic centimeters (milliliters = mils.), and all weights in Grammes. 2. Arabic numerals should be used in all cases; and the numbers should be placed before the abbreviations. Thus, 2.5 Gm., 160 Cc, 5 cm., etc. 3. Since so much depends upon the proper placing and the proper reading of the decimal point, special precautions should be observed to avoid a chance for a mistake. It is a common practice to draw a vertical line in the proper place on the prescription blank, and to write whole numbers of 14 Pharmaceutical Arithmetic mils (c.c.) or of Grammes to the left of the line, and the fractions beyond the line. Thus the decimal point is done away with. Example. — Quin. Bulph. 2 Ferri phos. 8 75 Potass, citr. 8 75 Syr. calc. lactophos. 125 Aquae 15 Elix. Arom. q. s. ad 500 Since it is customary in this country to prescribe all liquids by measure, and all solid? by weight, there is no difficulty in determining which numbers in a prescription stand for mils (c.c), and which for Grammes. Quinine sulphate being a solid, the "2" following this item in the prescription given, stands for 2 Gm. But the " 125," ex- pressing the quantity of the syr. calcium lactophosphate, obviously means 125 mils; for the article is a liquid, and hence is to be measured. If the vertical line is not made use of, every decimal fraction in the prescription should be written with a cipher before the decimal point. Thus, 0.75 Gm. in place of .75 Gm. ; 0.1 mil in place of .1 mil. This precaution makes it less probable that imperfections in the paper, or accidental markings, be mistaken for decimal points. The older prac- tice of using a comma (,), in place of a point, is also to be recommended. However, the best safe guard is in the use of the vertical decimal line. Reading Metric Quantities in Prescriptions. — A whole number expressing a metric quantity is always read as a simple denominate number, and not as a compound number ; that is, only one unit is used, no matter how large the num- ber. Thus, 3450 Gm. is read — three thousand four hundred and fifty Grammes; — not three Kilogrammes, four Hecto- grammes, and five Dekagrammes, which would be unnec- essarily cumbersome. 6565 Cc. is read — Six thousand five The Metric System i$ hundred and sixty-five cubic centimeters — ^not six Liters, etc., etc. Fractions of mils (c.c.) are read as tenths or hun- dredths, as the case may be. Fractions of Grammes are read as so many of the lowest unit in the fraction, it being remembered that the first decimal place belongs to deci- grammes, the second to centigrammes and the third to milligrammes. According to this rule 0.5 Gm. would be five decigrammes; 0.55 Gm., fifty-five centigrammes; 0.555 Gm., five hundred and fifty-five milligrammes. In case of a fourth decimal, this is read as so many tenths of milli- grammes ; and .3575 Gm., as three hundred and fifty-seyen and five-tenths milligrammes. Some pharmacists ignore the units, decigramme and centigramme, and read all fractions of grammes as milli- grammes. Thus 0.5 Gm. would be five hundred milli- grammes, and 0.55 Gm., five hundred and fifty milli- grammes. In dictating prescriptions by telephone, especially if there are mixed decimals to dictate, the plan of reading the numerals from left to right (in place of the whole numbers) should be adopted. The pharmacist can then copy the numerals as fast as read, and need not hold the entire num- ber in his mind — a task which is vei^^ trying, and may lead to error. To illustrate: the prescription — 9 Arsenous acid .065 Ext. opium .48 Quinine sulphate 8.75 Make 65 pills. Should be telephoned — Arsenous acid — point — nought — six — ^five, Ext. opium— point — four — eight. Quinine sulphate — eight — point — seven — ^five. Make 65 pills. Of course Grammes would be understood. l6 Phannaceutkal Arithmetic. Metric Prototypes. As has been stated, the metric system was based on the ivt^ootu °^ the earth's quadrant A bar of platinum was con- structed of this length, and was deposited in the French Archives, to serve as a prototype or model for the meter measures intended for actual use. There was also constructed, and deposited In the French Ar- chives, a weight of platinum of such size as to counterpoise in vacuo one cubic decimeter of water at its greatest density. This weight constituted the fundamental standard of mass, and was to serve as a prototype or model for the Kilogramme weights (and indirectly for the other metric weights) intended for actual use. The metric system was adopted by France in 1795. The pro- totypes referred to became the legal reference standards of France shortly after the adoption of the system. In 1875 an International Metric Convention was held In Paris, the United States participating. This convention called into being an International Bureau of Weights and Measures, the first duty of which bureau consisted in preparing an inter- national standard Meter bar, and an international standard Kilo- gramme weight, and duplicates of these for the seventeen coun- tries which had contributed to the support of this bureau. It was decided that the international prototypes and their duplicates should be made to equal (the Metric bars in length, and the Kilogramme weights in mass) the prototypes of the French Archives; and should be constructed of an alloy of 90 parts of platinum and 10 parts of iridium, a composition selected on account of its hardness and its resistance to the action of atmospheric gases. Of the 30 Meter bars and 31 Kilogramme weights made, one Meter bar and one Kilogramme weight were selected as inter- national prototypes and placed in the charge of the international bureau. The others were distributed among the countries which had called the bureau into existence. The distribution was ef- fected by lot, the United States drawing Meters No. 21 and No. 27, and Kilogrammes No. 4 and No. 20. Meter No. 27 and Kilogramme No. 20 were selected as our National Prototypes, and are carefully preserved in the United States Office of Weights and Measures, which is a branch of the Treasury Department. The Meter No. 21 and the Kilogramme No, 4 are used as working standards. While the 30 Meters and the 31 Kilogrammes were made &a nearly alike as it was possible to make them, it was found that they were not precisely alike. But the difference between the International Prototypes and the several national prototypes was In each case determined with the utmost care. Thus our Na- The Metric System I7 tional Kilogramme (No. 20) was found to be .00003 Gm. too light; and this fact must be taken cognizance of in the trtandard- izations of other weights. The two National Prototypes — the Meter No. 27 and the Kilo- gramme No. 20 — are at the present time the ultimate reference standards for all weights and measures in use in the United States, with the sole exception of the weights used in the U. S. mints, which weights are by law still referable to the standard troy pound received from England in 1827. Advice to the Student. The student is advised to acquaint himself with the metric length units by the use of a metric rule; with the volume units by means of measuring vessels graduated in mils (c.c.) and Liters; with the units of mass, by means of a set of metric weights. The Gramme and the mil (c.c.) should be known to the pharmacist as the pound and pint are known to the grocer — from actual use of the weights and measures. It is a mistake to think of the Liter as equivalent to so many fluid ounces, and of the Gramme as so many grains. Such equivalents have their uses; but the student must become familiar with the metric units by using measures and weights ; — not through equivalents. To learn French properly, one must learn to think in French — not to think in English and then to translate. In like man- ner, one must learn to think in metric units — to guess at the diameter of a filter in centimeters, the capacity of a flask in cubic centimeters, the weight of a mass of quinine in Grammes. As long as the metric units are abstract ideas, the metric system will appear intricate and hard to master ; but when these units find expression in something material — in graduated measures and sets of weights — all difficulty disappears at once. Problems and Exercises I. Read the following metric expressions: (a.) 6560 c.c; (b.) .02 L. ; (c.) 2 L. expressed in mils; (d.) 1.64S Gm.; (e.) 65 mg.; (f.) 6 eg.; (g.) 6 dg.; (h.) 6 Kg.; (i.) l8 Pharmaceutical Arithmetic 6 Kilo.; (j.) 17 M.; (k.) 7 cm.; (1.) 700 mm.; (m.) 6789.123 M.; (n.) 6789.1 c.c; (o.) 6789.123 Gm. 2. Write in numbers, followed by the proper abbrevia- tions:— (a.) six milligrammes; (b.) sixteen centigrammes; (c.) six tenths milligrammes; (d.) six Liters; (e.) six and six tenths milliliters; (f.) six and six tenths cubic centi- meters; (g.) six hundred milligrammes. 3. By moving the decimal point read — (a.) 6000 c.c as L. ; (b.) 1455 Gm. as Kg.; (c.) 414 mm. as M. ; (d.) .45 Gm. as eg.; (e.) .45 Gm. as mg. ; (f.) 40 mg. as Gm. ; (g.) 4 dg. as mg. 4. Supply the proper abbreviations in the following formula: — Guaiac 12 5 Potass, carbonate 6 Pimenta 3 Pumice 6 Alcohol 43 5 Water 43 5 Diluted alcohol q. s. ad 100 Addition. 5. Add: 2Kg., 43 Gm., 42 eg., 425 mg. Solution. — 2 Kg. = 2X1000 = 2000 Gm. 43 Gm. = 43 ** 42 eg. = 42-f-ioo = .42 " 425 mg. = 425-f-iooo = .425 " 2043.845 Gm. Suggestion. — ^To simplify addition and subtraction, ex- press all lengths in Meters, all volumes in mils (c.c), all weights in Grammes. Then weights may be added to weights, volumes to volumes, and lengths to lengths, in the same manner in which abstract numbers are added. The Metric System 19 6. Add: 10 L., 100 c.c, 5.6 L., 34 c.c. Ans. 15703-4 C.C. 7. Add: 2 Kg., 4 Dg., 500 Gm., 32 mg., Ans. 2540.032 Gm. 8. Add : 2 cm., 25 cm., 5 M., 500 mm., 13 Km. 9. Add: 5 eg., 50.5 mg., 5.5 Gm. 10. Add: 6 Kg., 20 Hg., 200 Dg., 2145 Gm. Ans. 12145 Gm. Subtraction. 11. Subtract 345 c.c. from 5 L. Solution. — 5 L. = 5000 c.c; and 5000 CC- — 345 CC = 4655 c.c. 12. Subtract 345 mg. from 140.002 Gm. Solution. — 345 mg. = .345 Gm.; and 140.002 Gm.— » .345 Gm. = 139.657 Gm. Subtract : 13. I cm. from 500 mm. Ans. 490 mm. 14. 2 L. from 6050 c.c. Ans. 4050 c.c 15- 3-5 c.c from i L. Ans. 996.5 c.c. 16. 1.4 mg. from 10 Gm. Ans. 9.9986 Gm.- 17. A druggist making solution of lead subacetate finds that his evaporating dish plus contents weighs 757 Gm. The dish itself weighs 289.02 Gm. How much water must be added to make the contents weigh 500 Gm? Ans. 32.02 Gm. MuivTlPLI CATION. Remarks. — The product is always in the same unit [denomination] as the multiplicand. If mg. are multiplied, the product will be mg. ; if Gm. are multiplied, the product will be Gm. ; etc. If a length, volume, or weight is expressed in a com- pound denominate number — ^that is, if several units are used in expressing it — reduce to a simple denominate number be- fore multiplying. 20 Pharmaceutical Arithmetic Remember that the decimal places in the product must equal the total number of decimal places for both factors. That is, if the multiplicand has 3 decimals, and the multiplier 4, the product must have [3+4 =] 7. 18. Multiply 5.045 Gm. by .004. Operation.— <,.04^ Gm. X .004 .0201 8jz( Gm. 19. Multiply 345 mg. by 856. Ans. 295320 mg. = 295.320 Gm. 20. Multiply 50.5 c.c. by 7.058. Ans. 356.429 c.c. 21. A prescription is received for 48 capsules, each to contain . 195 Gm. of mercurial mass, . 12 Gm. of comp. ext. colocynth, .15 Gm. of ext. gentian. How much must be weighed out of each ingredient? Division. Remarks. — The quotient is always in the same unit [denomination] as the dividend. Remember that the quotient must have as many decimal places as those in the dividend exceed those in the divisor. If the decimal places in the divisor exceed those in the dividend, add ciphers to the latter until the decimal places equal those in the divisor. When a denominate number expressing volume or weight is to be divided by another denominate number also expressing volume or weight, both must be in the same unit [denomination]. See prob. No. 27. 22. Divide 95.789 Gm. by 7.42. 23. Divide 74.2 mg. by .02. Ans. 3710 mg. 24. Divide .04 Gm. by 64. 25. Divide 6 Kg. by 8. Ans. 750 Gm. The Metric System 21 Suggestion. — Since the number in the dividend in prob- lem No. 25 is smaller than the number in the divisor, reduce to Gm. before dividing. Remember that the quotient must have .as many decimal places as those in the dividend exceed those in the divisor. 26. Divide 30 Kg. by 8. Ans. 3750 Gm. Suggestion. — Since the number in the dividend cannot be divided by 8 without a remainder, it is simpler to reduce 30 Kg. to Gm. before dividing. It is not customary to ex- press metric quantities by compound denominate numbers. 27. A Seidlitz powder contains 7.75 Gm. of Rochelle salt. How many powders could be made from 2 Kg. of the salt? Ans. 258. Miscellaneous Problems. 28. A druggist receives four lumps of opium, weigh- ing respectively (i.) i Kg. 20 Gm., (2.) 5 Hg. 3 Dg., (3.) 650.35 Gm., (4.) 205 Gm. 15 mg. How many Gm. of opium does he receive? Ans. 2405.365 Gm. 29. Suppose opium were worth $13.50 per Kilo. What would the above consignment cost? Ans. $32.47. 30. A certain elixir is to contain .175 mg. of strych- nine in each c.c. How much strychnine must be weighed out for 3 L. of the elixir ? Ans. 525 mg. 31. A certain prescription calls for 75 pills, each to con- tain 200 mg. of quinine sulphate. How much of the latter must be weighed out? Ans. 15 Gm, 32. A prescription calls for 5.1 Gm. of pancreatin, to be mixed with 20.4 Gm. of sodium bicarbonate, and the mixture to be divided into powders weighing 1.7 Gm. each (amount for peptonizing 500 c.c. of cow's milk). How many powders will the mixture make? Ans. 15. 06 33. B Strych. sulpli. Quin. sulph. 16 Ferri phos. 10 Mix and divide into 85 pills. 75 22 Pharmaceutical Arithmetic How much strychnine sulphate, how much quinine sul- phate, and how much ferric phosphate in each pill? 34. A pharmacist extracts 7 headache tablets with chloroform, for the purpose of dissolving the caffeine pres- ent. He evaporates the chloroformic solution to dryness in a beaker, and finds that the residue and the beaker together weigh 11.005 Gm. The beaker itself weighs 10.38 Gm. How much caffeine in each tablet? Ans. .08928 Gm., which read, eighty-nine and twenty-eight hundredths milligrammes. B. English Systems Customary in the United States. While the metric system has had legal sanction since 1866, has since been adopted for all scientific work, and is the official system of our pharmacopoeia, it has not as yet displaced the cumbersome English Systems in our every-day business transactions. Lumber is still sold by the foot, cloth by the yard, coal oil by the gallon, and meat by the avoir- dupois pound. For prescription work the use of the old English troy weight — now obsolete in England — is stiil quite general in the United States. I. Linear Measure. The English linear measure, with its units of inch, foot, yard, etc., was originally derived from the average length of the barley corn. According to an English law of the year 1324, "three barley corns, round and dry, shall make an inch, and twelve inches a foot." But at the present time the inch is defined in England as the-j^.-j-J^yg^of the length of the second's pendulum, which latter is therefore the natural standard of the English linear measure, and through the same, of all the English systems of measure and weight.* •The leng^th of a pendulum determines the time of its vibrations. The shorter the pendulum, the shorter are its vibrations; the longer the pendulum, the longer and slower are its vibrations. A pendulum 39.13929 inches in length will always make one complete vibration ("swing-swang") per second, provided the pendulum is placed in a vacuum, the temperature is 62 degrees F., the altitude is that of the sea level, and the latitude, that of London. Conversely, a pen- dulum, beating seconds of time under the aforementioned conditions is bound to be just 39.13929 inches ifa length. And since a second's pendulum can be constructed as often as necessary, it serves admirably as a natural standard,— i. e., as a check upon the material standards in use. English and American Systems 23 The lack of simplicity in the relationship between the inch and the length of the second's pendulum — ^the latter being 39.13929 times that of the former — is readily account- ed for by the fact that the length of the inch had become fixed by custom and by law, before the second's pendulum was selected as the natural standard. In the United States the inch is now defined as equiva- lent to 25.4001 millimeters. The yard is defined as If^ Meters. All our measures and weights are now defined in metric terms, thus making the metric prototypes the standards for all the weights and measures in use in the United States, excepting weights used in the U. S. Mint, which are still regulated, in accordance with a law of Congress, by the standard troy pound, received from England in 1827. To repeat: — the American and the British yard are identical in length ; but the former has lor its prototype the Meter bar No. 27 (see page 17) and for its natural standard, the earth's quadrant ; while the British yard is referable to a certain bar, with a gold peg near each end, a bar deposited in the office of the Exchequer, London; — and the natural standard for the British yard is the second's pendulum. Table a. 12 inches (in.) = i foot (ft.) 3 feet = I yard (yd.) [Higher units are never used in pharmacy.] Table B. Yard Feet Inches I = 3 = 36 I = 12 24 Pharmaceutical Arithmetic 2. Apothecaries' Fluid Measure (U. S. Wine Measure). The primary unit of the U. S. Wine Measure is the gallon, which is defined by the National Office of Weights and Measures as the volume of 3785.33 mils or 3785.33 Gm. of pure water. Prior to 1893 the gallon was officially defined as 231 cubic inches, which shows its relation to the English Linear measure. The Wine Measure, as used in this country, with its units of gallon, quart, and pint*, was brought from England, in which country, however, it has long since (1826) been superseded by the Imperial Measure. The Apothecaries' Fluid Measure is the Wine Measure minus the quart unit, and with the addition of three units smaller than the pint. The table, giving symbols and abbreviations, is as fol- lows: Table A. 60 minims (tt],) = i fluid drachm (f3) 8 fluid drachms = i fluid ounce (f^) 16 fluid ounces = i pint (O.) 8 pints = I gallon (Cong.) O. is the abbreviation for the Latin word Octarius; Cong., for the Latin word Congous. Table B. Gallon Pints Fluid Ounces fluid Drachms Minims I = 8 = 128 = 1024 = 61440 I = 16 = 128 = 7680 ' = M 00 11 II 480 60 *T^e gill, which is ^ pint, is obsolete, but is occasional!; met with in old ftmily recipes. English and American Systems 2$ Apothecaries' Measure in Prescription Writing. — Rom- an numerals are used in connection with this system. It is customary to place the symbol or abbreviation before the number — not after it, as in case of metric prescriptions. Example : H. TInct. ferri chloridi f 3 v j Acidi phosphoric! diluti f 3 iij Spiritus limonis f 3 ij Syrupl q. s. ad f S vj Notes. — The dot over the numeral I Is used as a precaution against ambiguity which might be occasioned by carelessness in writing. For example, U might stand for II or for V; but if the dots are systematically used, their absence would prove the character to be a V, while their presence would prove it to be a XL In written prescriptions the terminal i is usually replaced by j as a further precaution against misinterpretation. Many physicians omit the f, writing 3 for f3, and 5 for f^. Since liquids are measured, and solids weighed, the 5 and 5 stand respectively for f3 and for fg in case of liquids, and for the weight units in case of solids. Roman Numbers. Numerals. — The Roman numerals are: I = 1; V = 5;X = io;L = 5o;C = lOO ; D = 500 ; M = 1000. [ss., from semis, Latin for half, is used in prescriptions.] Rules for combining the numerals. — 1. Repeating a numeral repeats its value. Thus, III = 3 ; XX = 20 ; CCC = 300 ; etc. But since V, L, and D if doubled would give the equivalent of X, C, and M re- spectively, these three first-mentioned are never doubled. In place of W, X is used ; in place of LL, C is used ; etc. 2. Numerals placed after a numeral of higher value are added to the latter. Thus, VI = 5 + 1, or 6; XI = 11; XV = 15; XVII r= 17; CLXVI — 166. 26 'Pharmaceutical Arithmetic 3. A numeral placed before a numeral of higher value is subtracted from the latter.* Thus, IV = 5 — I, or 4; IX = 9; XL = 40; XC = 90; XLVI = 50 — 10 + 5 4- I, or 46; XCVIII = 98. 4. A numeral placed between two higher numerals is always read in connection with the one which follows, and according to Rule 3, is subtracted. Thus, XIV = 14; XXXIX = 39; LIV = 54; XLIV = 44; MDCCCLXXXIV = 1884; MDCCCVC = 1895. Problems and Exercises. 36. Write the following expressions of volume as they should be written in a prescription : — (a) 45 minims, (b) 6 fluid drachms, (c) 14 fluid ounces, (d) 3 pints, (e) i gal- lon, (f) 2^2 fluid ounces, (g) 2 fluid drachms and 30 minims. REDUCTION Descending. 37. Reduce 2 Cong, i O. 2 f 5 i f 3 20 TT to tn,. Suggestion.— Reduce the 2 gallons to pints by multiplying by 8; to the product add the 1 pint given; reduce the' total number of pints to f5 by multiplying by 16; add the 2 tS given; reduce fS to f3 by multiplying by 8; add the 1 fS given; reduce f 3 to Ti^ by multiplying by 60; finally add the 20 Tit given. 38. Reduce ^ Cong, to f 5. Solution. 3«V X 8 X 16 = ^i'^ f 5 = 22 Iff 5. Since our meaa ures are not graduated in fractions, it is customary to reduce fractions to lower units. — J*f5 = HX 8=fM3= 4 iff 3. If f 3 = H X 60 = W ^ = 42 A ^a. Then A Cong. = 22 f § 4 f 3 42 3»y TH, 39. Reduce 3 O. 5 f 5 to TH.. Ans. 25440 in,. Reduce 6 Cong, to f 3. Ans. 768 f 5. -^ 40. Reduce 3f f 5 to in,. Ans. 1728 HI. •But only one numeral can be so subtracted. English and American Systems 2y RieDucTioN Ascending. 41. How many fluid ounces in 6000 minims. Solution. — In one fluid ounce ttiere are 480 minims (See Table B) ; hence there are as many fluid ounces in 6000 minims as 480 is contained in 6000. Ans. 12 1^ fj. 42. A physician wishes to write a prescription for 64 one-fluid-drachm-doses of syrup of hypophosphites. What volume should he prescribe? 43. How should the volume of 8000 TT\. be expressed in order that it might be measured with the customary measures? Ans. i pint 5 f3 20 TT].. 44. Convert 666 fluid ounces into higher units — pints and gallons. Solution. — Since there are 16 fluid ounces in a pint, 666 -j- 16 = number of pints = 41 pints and 10 fluid ounces over. There being 8 pints in a gallon, 41 -=- 8 = number of gallons = 5 gal- lons, and 1 pint over. The answer, therefore, is 5 gallons 1 pint 10 fluid ounces. Remark. — ^Volumes in apothecaries' measures are ex- pressed by compound denominate numbers rather than by simple denominate numbers with fractions — unless the fraction be ^, in which case it may be used. Addition. 45. Add the following volumes: — 6 gallons; 4 gallons 3 pints 7 fluid ounces 5 fluid drachms ; 3 pints 8 fluid ounces 4 fluid drachms. Suggestion. — Arrange the denominate numbers under each other so that each unit will form a separate column. Add the column of the smallest unit first. If the sum is larger than the equivalent of the next higher unit, reduce to that unit, but enter the remainder under the column added. Proceed in this way until all columns have been added. Operation. — gallons 6 pints fl. ounces fl. drachms 4 • • 3 I 3 8 5 4 10 28 Pharmaceutical Arithmetic 46. Add: 15 gallons 5 fl. ounces. I gallon 7 pints 12 fl. ounces. 5 fl. ounces 6 fl. drachms. 10 fl. ounces 4 fl. drachms. Subtraction. 47. Subtract i pint 12 fl. ounces from 5 gallons. Suggestion. — Arrange subtrahend under minuend so that each unit of the former comes under the same unit in the latter; — i. e., gallons under gallons, pints under pints, etc. Subtract from right to left, that is, the lowest unit first, then the next higher, and so on. If then, in any column except the last one to the left, the subtrahend is found to exceed the minuend, one of the next higher unit in the minuend must be reduced. Operation. — gallons pints fl. ounces. Gallons pints fl. ounces 5 4 7 16 — I 12 = — I 12 46 4 4 6 4 - 48. Subtract fi f 5 from li f %. Operation: 15 X 4 = 60. Hence 60 is the least common multiple. Ans. H f 5 Note. — ^But this answer would require reduction to lower units in most practical problems. Another process would be to first reduce the subtrahend and minuend to lower units; then to subtract in the usual way. 49. A druggist taps a 36 gallon barrel of alcohol, mak- ing an air hole which he forgets to close. By keeping an account of the alcohol drawn oflF, he finds that the barrel supplied 31 gallons 4 pints 7 fl. ounces. How much was lost by evaporation? Multiplication. 50. Multiply 5 pints 6 fl. oimces 2 fl. drachma 40 minims by 7. English and American Systems 29 Suggestion. — ^Multiply smallest unit fxrst, then the next high- er, and so on. In each case reduce the product to tne next higher unit when that is possible, and write only the remainder under the unit multiplied. Operation. — 0 f5 f3 m 5 6 2 40 X 7 Ans. 4 Cong. 5O. 12 f^ 2f3 40Trt 40 ni X 7 = 280 Til = 4 f 3 and 40 iTt 2 f 3 X 7 = 14 f3 ; 14 f3 + 4 f3 — 18 f3 = 2 fg and 2 f3 6 f 5 X 7 = 42 f S ; 42 f 5 + 2 f B = 44 f S = 2 O. and 12 f S 5 O. X 7 = 35 O. ;35 0.+20.= 37 O. = 4 Cong, and 5 O. 51. Multiply f m by 60. Operation: f X 60 = -"^ = 17+ [m] 52. Multiply ^ f 5 by f. operation: f X i ■= i| f !• 53. Multiply 50 ni by \. Operation: 50 X i = Y X i = V" = 10 [^] Note. — It will be seen that multiplication by a fraction la virtually a method of accomplishing division, 54. Multiply 600 Til, by 20. Reduce product to higher units. Ans. i pint 9 fl. ounces. 55. How much creosote is required to make 4 gross of 5 minim capsules ? Division. 56. Divide 10 gallons 5 pints 4 fl. ounces by 6. Suggestion. — Divide from left to right, beginning with the highest unit. In case there is an indivisible remainder, reduce this to the next lower unit, and add to the number of that unit be- fore the same is divided. Operation. — Cong. O. 1% 6) 10 5 4 163 and 2f340Tn. 30 Pharmaceutical Arithmetic lo Con. -^- 6 = I Cong, and J over ; 4 Cong. = 32 O. ; 32 O. + 5 O. = 37 O. 37 O. -J- 6 = 6 O. and i over ; i O. = 16 f5; 16 f5 + 4 f5 = 20 f5. 20 f5 ^ 6 — 3 f§ and 2 f5 over. 2 £5 = 16 f3 ; 16 3 -^ 6 = 2 f3 and 4 f3 over. 4 f 3 = 240 HI ; 240 TTi. -^- 6 = 40 TTt- 57. Divide i m by 5. operation: l-^f = iXi = iV[Ta] 58. Divide i m by i. Operation: i--i=iXf = i = li[Ta] Note. — ^Just as multiplication by a fraction is a method of effecting division, so a division by a fraction is a method of multiplication. 59. Divide 5 gallons 2 pints 2 fl. ounces by 8. 60. (a.) 3 O. -T- 24; (b.) 4 Cong. --- 15; (c.)' 4 il 61. With 8 fl. ounces 2 fl. drachms of oil of sandalwood in stock, how many 4 minim capsules could be filled? Ans. 990. 62. (a.) How many 2 fl. drachm doses in an 8 fl. ounce mixture? (b.) How many 15 minim doses in i pint? 3. Im PERI At Measure. This is the fluid measure of Great Britain. It has been in use since 1826, when it superseded the wine measure, which is now the customary fluid measure of the United States. The Imperial Measure is superior to the measure it dis- places inasmuch as it bears a simple relationship to the sys- tem of weight official in that country, namely the avoirdu- pois weight (page 32). The Imperial gallon, which is the primary unit, is de- fined as the volume of 10 av. pounds of pure water, at 62 °F., and at a barometric pressure of 30 inches. The Imperial fluid ounce is the volume of i av. ounce of pure water, at 62 °F., and at a barometric pressure of 30 inches.* *See foot note i>afire 12. English and American Systems $1 Table A. 60 minims (Til,) = 1 fluid drachm (fl. drm.) 8 fluid drachms = 1 fluid ounce (fl. oz.) 20 fluid ounces = 1 pint (0.) 8 pints = 1 gallon (C.) As explained on page 41, the first three Imperial units are smaller, and the pint and gallon larger than the corresponding units of the apothecaries' measure. Table B. Gallon. Pints Fl. Ounces. Fl. Drachms. Minims. 1 = 8= 160 = 1280 = 76800 1 = 20 = 160 = 9600 1 := 8 = 480 1 = 60 Problems in Imperial Measure. 63. (a.) How many fl. oz. in 5 C? (b) Ho-w many fl. drm. In 5 O.*? (c.) How many tn, in 2 fl. oz.? 64. Reduce 4 C. 3 O. 14 fl. oz. 2 fl. drm. to TT},. Ans. 342840 m.. 65. Convert 5000 IT], to higher units— fl. drm., fl. oz., etc. 66. Add 6 C. 4 O. 12 fl. oz.; 7 O. 10 fl. oz.; 3 fl. oz. 40TTI. 67. Subtract 5 fl. oz. 4 fl. drm. from 5 C. Ans. 4 C. 7 O. 14 fl. oz. 4 fl. drm. 68. Multiply 2 fl. oz. 4 fl. drm. 40TI1, by 24. Ans. '3 O. 2 fl. oz. 69. Divide 3 O. by 48. Ans. 1 fl. oz. 2 fl. drm. 70. Divide 1 C. 2 O. 2 fl. oz. 3 fl. drm. by 16. 71. A Canadian druggist receives a 4 pint bottle of Eaton's syrup from a British drug firm. How many 4 fl. oz. bottles can he fill? Ans. 20. 4. Avoirdupois Weight. All weighable merchandise, except jewelry and precious stonesf, are generally bought and sold in this country by avoirdupois weight, — the metric weight not having been adopted as yet for every-day commercial transactions.ff *0 in this paragraph stands for Imperial pint; elsewhere for U. S. pint. Unless the contrary is expressly indicated, gallons, pints, fl. ounces, fl. drachms, and minims are in this country understood to be those of the U. 8. fluid measure. tSold by troy weight:— 24 grains = 1 pennyweight; 20 pennyweights = 1 ounce; 12 ounces ^ 1 pound. ttTo assist in popularizing the metric system, the firm of E. R. Squibb At Sons has for sereral years put up their products in packages of metric quantities! 32 Pharmaceutical Arithmetic It should be distinctly understood that in the United States all drugs and chemicals which are not bought and sold by metric weight, are bought and sold by avoirdupois weight. The apothecaries' weight is never used in commer- cial transactions, not even in case of the most costly plant principles. An ounce bottle of morphine, or of quinine, con- tains 437^ grains ; an eighth ounce bottle of aconitine con- tains j/s of 437^ grains. The avoirdupois system of weight was introduced into England about 6oo years ago. It is now known in that country as the Imperial weight, and is used in prescription compounding as well as for buying and selling. It is the official weight of the British Pharmacopoeia. In the United States, the units of the avoirdupois weight are officially defined in terms of metric units. The smallest unit, the grain, is defined as equivalent to 64.7989 milli- grammes. Table A. : 437/^ grains (gr.) = i ounce (oz.)' 16 ounces = i pound (lb.) Table B. Pound. Ounces. Grains. I = 16 = 7000 I = 437^ Note. — ^The avoirdupois weight Includes a unit called aVi. drachm, which is ^ of an ounce, and is equivalent to 27 fi grains. But this unit is seldom used, it being customary to divide the ounce into halves, quarters, eighths, etc. Problems in Avoirdupois Weight. 72, (a.) Reduce 16 av. lb. to oz. ; (b.) 4 lb. to gr. 73. (a.) How many grains in }i oz. of cocaine? (b.) How many grains in }^ oz. bottle of atropine ? English and American Systems 33 74. How many 3 gr. capsules cx)uld be filled from a I oz. bottle of quinine ? 75. How many grains of cocaine hydrochloride would be left in a ^ oz. bottle after removing 30 gr. ? Ans. 24 J^gr* ^6. How many }i gr, morphine sulphate tablet tritu- rates could be made from the contents of a ^ oz. bottle ? y;. (a.) Reduce 50000 gr. to lb. and oz. ; (b) 35000 gr. to lb. ; (c.) 3000 gr. to oz. 78. A druggist desires to put up 5 gross of xo gr. doses of Compound Acetanilid Powder (N. F.) How much of the latter must be made for the purpose? Ans. i lb. 200 gr. 79. 15 lb. 6 oz. 4- 6 lb. 10 oz. 100 gr. + ii oz. 400 gr. = ? Ans. 22 lb. 12 oz. 62^ gr. 80. From a 5 lb. can of opium the following quantities have been removed: — 3 oz., i^ oz., 1.64 oz., 175 gr. How much remains? 81. The British Pharmacopoeia gives the following formula for Gregory's powder : rhubarb, 2 oz. ; light mag- nesia, 6 oz. ; ginger i oz. How many 20 gr. doses would this make ? Ans. 196 -\- doses. 5. Apothecaries' Weight. This is a modification of the old English troy weight (See note page 31 and note page 34) the two being the same except in the units between the ounce and the grain. The apothecaries' weight has never received legal recog- nition in this country; but since the apothecaries' grain is identical with the avoirdupois grain, and the latter is now defined in metric units based on the National Prototype Kilogramme, this reference standard may serve also for the apothecaries' system. The smallest unit — ^the grain — was originally derived from the average weight of a grain of wheat. Later the weight of a certain volume of water [ i cu. in. of water, in vacuo, and at 62° F. = 252.75965 gr.] was used as the basis 34 Pharmaceutical Arithmetic of weight units in England; but the name grain for the smallest unit of weight has been retained to this day.* When the Pharmacopoeia of London was compiled (in 1618), the troy (apothecaries') weight was decided upon as the weight to be used in compounding medicines. In Eng- land it has since been superseded by the Imperial weight (avoirdupois) ; but in this country the apothecaries' weight is so firmly established, that the inconveniences incident to the use of one system of weight for compounding, and a different system for buying and selling, will, no doubt, con- tinue to trouble the pharmacist for many years to come. Table A. 20 grains (gr.) = i scruple (9) 3 scruples = i drachm (3) 8 drachms = l ounce ( 3) 12 ounces = i pound (lb) Table B. ' !b 5 3 9 gr. I = 12 = 96 = 288 = 5760 I = 8 = 24 = 480 I = 3 = 60 I = 20 Notes. — (1.) The apoth. pound Is seldom used in pharmacy. (2.) The troy, grain, apoth. grain, and the avoirdupois grain are identical. (3.) The apoth. ounce, and the troy oimce are the same, but the av. oimce is 42% gr. smaller. (4.) The apoth. pound and the troy pound are the same, but the av. pound is 1240 gr. larger. •In 1266, in the reijrn of Henry m, the following law was enacted :—" An English silver penny, called a sterling, round and without clipping, shall weierh 82 grains of wheat, well dried and gathered out of the middle of the ear; and twenty pence (pennyweights) do make an ounce, and twelve ounces a pound." Subsequently, in the reign of Henry VII, about 1497, the pennyweight was divided into 24 parts, called grains, in place of into 82; thus creating the system of troy weight as used to this day. English and American Systems 35 Apothecaries' Weight in Prescriptions. — Roman num- erals are used ; and the numbers always follow the symbols or abbreviations. In case of fractions, with the exception of J^, expressed by ss , which is the abbreviation for semis, Latin for one-half, Arabic numerals are used. Example : I^ Acidi arsenosi gr. | Strychninae sulphatis gr. ^ Quininae sulphatis 9 Ij Ext. gentianae 3 ss Ft. pil. no. XX. Problems in Apothecaries' Weight. 82. Reduce 4 § 2 3 i 9 to gr. Ans. 2060 gr. 83. Reduce 7000 gr. to higher units — 9, 3, § and tb. Ans. itb. 2§432 9. 84. (a.) Reduce 1000 gr. to 3 ; (b.) 7000 gr. to lb. ; Co.) 2000 gr. to S; (d.) 437^ gr. to §. 85. A certain formula for a horse powder is as follows : Black antimony S viij Sulphur Svj Saltpetre 3vj Resin 3 V Capsicum 3 iij 3j How much will the formula make ? 86. Subtract 2 3 1 9 12 gr.from 8 §. Ans.75 53 1 9 8gr. 87. The following prescription is to be filled, making eight (8) times the quantity: IJ Ext. ergotae 3 j Ferri sulphatis 3 jss Ext. nucis vomicae gr. viij Hydrarg. chlor. corr. gr. ss Make 30 pills. How much of each ingredient must be weighed out? 36 Pharmaceutical Arithmetic 88. Divide i 3 i9 lo gr. by 8; (b.) by 8o; (c.) b^ 8oo; (d.) by .08. 89. Divide I B 4 3 by ^ ; (b.) by t\; (c.) by 1.034 (d.) by .0055. Ans. (d.) 272 5 5 3 2 9 9 gr. 90. B Zinci phosphidi 3ij Ext. nucis vomicae 3 j Ext. gentianae 3 iv Make 80 pills. How much of each ingredient in each pill ? 91. 3 Camphorae gr. Ixv Ammonii carbonatis gr. xlviij Pulv. opii gr. ix Mix and divide into 3 gr. doses. How many will it make ? 92. 9 Rhei Sij Aloes S iss Myrrhae 3j Saponis 3j Olei menthae piperitae gr. viij Glucosi 3 j 3 j Mix and divide into 4 g^. pills. How many should be made? Ans. 104 pills, and over. 93. 9 Make 48 pills each to contain Atropine ^^ Morphine ■J Excipient q. s. to make a 2 gfr. pill. Note. — Since 4f gr. of atropine cannot be weighed, pro- ceed as follows : — Weigh out i gr. of atropine and mix with 74 gr. of some inert powder, thus making 75 gr. of mixture, of which each gr. represents -^ gr. of atropine. Now weigh out 48 grains of this mixture — the amount containing 41 gr. of atropine. This rule can also be applied in case of solutions. Sup- English and American Systems 37 pose Yhs S^' of strychnine sulphate is to be contained in each fluid drachm of a 4 fluid ounce preparation. In 4 fl. ounces there are 32 fluid drachms ; hence i%\ gr. of strychnine sul- phate is required — an amount which cannot be weighed. Proceed thus : — weigh out l gr. of strychnine sulphate, and dissolve in enough water to make 100 minims. Each minim of this solution contains Yhf 8J- of strychnine sulphate; therefore 32 minims should be used in the prescription. Review Questions. 1. Upon what object in nature is the primary unit of the metric system (the meter) based? 2. What serves in England as the "natural standard" for the yard, and thus indirectly for the units of capacity and of weight? 3. How is the yard officially defined in the United States? 4. Is our yard identical in length with the British yard ? 5. How many systems of weight are employed in pharmacy in this country? 6. Which two are used in compounding? 7. Which one in commercial transactions only? 8. Which pound is practically obsolete ? Which drachm is obsolete? 9. How many gr. in a troy ounce, an apoth. ounce, an Imp. ounce, an avoir, ounce? CHAPTER II. Relationship of Systems, I. Mettric and English Linear Measures. Official Equivalents. Approximate Equivalenta 1 M = 39.37 in* 1 M = 395^ in.- 1 in. = 25.4001 mm. 1 in. = 25 mm. 1 ft. = .304801 M. 1 ft. = .3 M. 1 yd.= .914402 M. 1 yd.= .9 M. Note. — All equivalents printed In bold faced type should be committed to memory. While but one connecting link is really necessary, much laborious figuring may be avoided by having several such links — just as in a city many steps may be saved by having more than one bridge across a river. On the other hand, a multiplicity of equivalents is apt to prove confusing to the student and has been carefully avoided in these pages. For a complete table of equivalents see the U. S. Pharmacopoeia. When to employ accurate equivalents, and when ap- proximate equivalents, must be decided in each case that comes up in practical pharmacy ; but as a general rule, ap- proximate equivalents are sufficiently accurate. Problems. Use Official Equivalents, 1. Reduce 69.325 M. to in. Solution. — If I. M. = 39.37 in., then 69.325 M. will equal 69.325 X 39-37 in- = 2729.3 in. 2. Reduce 69.325 in. to M. Solution. — If I M. = 39.37 in., then 69.325 in. will equal as many M. as 39.37 is contained in 69.325. 69-325 -^ 39-37 = 1-684 [M.] (Ans.) •According to Ca.pt, A. K. Clark, who made the determination for the Britldi Government, the equivalent is: 1 M = 39.37&i32 in. But the official equivalent given above is sufScientljr accurate for all practical purposes. Relationship of Systems 39 3. (a.) Reduce 6o>4 yd. to M. ; (b.) 54^ ft. to M. 4. Reduce 4 yd. 2 ft. 7^ in. to M. Suggestion. — Reduce first to in. ; then to M. 5. How many inches in i dm? in i cm? in i mm.? 6. (a.) Reduce 154 mm. to in.; (b.) 25 cm. to in. 7. A certain tube has an inside diameter of ^ in- Give this in mm. Ans. 4.76 mm. 8. The tube is 4 ft. 8 in. long. Give in M. Ans. 1.42 M. Use Approximate Equivalents. 9. Aconite is described in the U. S. P. as being 10 to 20 mm. thick, and from 50 to 75 mm. long. Give its dimen- sions in inches. Ans. f to ^ in. thick and 2 to 3 in. long. 10. Viburnum opulus is from i to 1.5 mm. thick, and about 30 cm. long. Give its dimensions in inches. 11. Swedish filtering paper is kept in the laboratory in the following diameters: 5.5 cm., 7 cm., 9 cm., 12.5 cm., 15 cm., 18.5 cm., 24 cm., 38.5 cm. Give these sizes in inches. 12. A druggist receives a prescription for a capsicum plaster 8 cm. by 12 cm. Give size in inches. Ans, 3^ by 4f in. 13. A druggist orders a 5 in. funnel (diameter across top) ; but the jobber sends a 12 cm. funnel. Is the one sent larger or smaller than the one ordered ? How much ? 2. Metric and Apothecaries' Fluid Measures. Official Equivalents. Approximate Equivalents. 1 C.C. (or mil) =16.23 ni. i c.c. (mil) c=i6TTt.t I L. = 33.814 f 5 IL. =33.8f5 If 5 =29.57 mils* If^ =30 mils I O. (pt.) =473.179 mils I O. (pt.) =473 mils I Cong. =3.78543 L. Memorize the equivalents printed in bold faced type. • More accurately, 29.5737 c.c. t In the U. S. P. doses this equivalent is further rounded off to 1 cc. = 15 tn,. 40 PJtarmaceutical Arithmetic For calculating doses, and for all pharmaceutical work not requiring very g^eat accuracy, the approximate equiva- lents may be used. In the U. S. P. i c.c. r= 15 Tl], is con- sidered sufficiently accurate for dose calculations. Problems. Use Official Equivalents. 14. Reduce 3 pints to c.c. Solution. — 3 pints = 48 f^ ; 48 X 29.57 = 1419.36. Hence 3 pints = 1419.36 c.c. 15. Reduce i Cong. 2 O. 6 £5 to c.c. Suggestion. — Reduce to f 5 ; then to c.c. 16. Reduce 93.29 c.c. to TH,. Ans. 15 14 Til,. 17. Reduce 9 ^ cc. to TT\.. Ans. 152 in.. 18. Reduce (a.) 300 ml. to f §; (b.) 3.78543 L. to pints. 19. Reduce 4.5 ml. to Til,. Ans. 73.035 TTL. 20. Reduce (a.) 40^ TT], to cc. ; (b.) 10 f § to cc. 21. How many f § will a 500 c.c fiask hold? 22. How many c.c. will a pint bottle hold? Ans. 473.179 cc Use Approximate Equivalents. 23. Express the following doses in metric measure : (a.) croton oil, i to 2 HI ; (b.) tr. belladonna, 5 to 15 "ni- (c.) infusion buchu, i to 2 f^; (d.) tr. cinchona, i f 3. 24. Express the following formula in apoth. measure : Ans. Castor oil 75 mils 2 f 5 4 f 3 Mucilage of acacia 37.5 " 1 f 5 2 f 3 Orange flower water 25 " 6 f 3 40 111 Cinnamon water 62.5 " 2 f 3 40 TTl Relationship of Systems 41 25. Express in metric measure the following formula for a cough medicine : Tr. opium fS vj Fl. ext. ipecac fS j Fl. ext. sanguinaria fS j Syr. squill tS iiJ Syr. wild cherry f 5 ix Syr. tar q. s. ad O.J 26. Express the following doses in apothecaries' measure: — sol. potass, arsenite, 0.3 c.c. ; sol. potass, citrate, 15 c.c; fl. ext. ipecac, 0.9 to 1.9 c.c; spt. glonoin, 0.06 to 0.12 c.c. ; tr. gelsemium, 0.6 to 1.25 c.c. ; comp. decoction of sarsaparilla, 120 to 180 c.c. ; inf. cinchona, 30 c.c. ; oil of eucalyptus, 0.6 to 0.9 c.c. ; syr. of tar, ^.';^ to 7.5 c.c. 27. A druggist receives four "metric prescriptions"; the first for a 60 c.c. solution, the second for a 120 c.c. lini- ment, the third for a 360 c.c. emulsion, the fourth for a 15 c.c. eye wash. State in fluid ounces the size of the bottle to be selected in each case. Note. — American-made prescription bottles are. commonly gauged in fluid ounces. 3. Apothecaries' and Imperiai, Measures. The Imperial measure differs from our apothecaries' measure in two respects: — the Imperial minim equals but 0.96014 TTL (U. S.) ; and the Imperial pint is divided into 20 Imperial ounces, while the pint (U. S.) contains only 16 fS (U. S.) The equivalents usually employed are: 1 Imperial minim (ni) = .96 TH, (U. S.) 1 Imperial fluid drachm (fl. dr.) = .96 f 3 (U. S.) 1 Imperial fluid ounce (fl. oz.) = .96 f5 (U. S.) 1 Imperial pint (O.) = 1.2 O. (U. S.) [accurately, 1.20017501]. I Imperial gallon (C.) = 1.2 Cong. (U. S.) [accurately, 1.20017501]. 28. Convert 6 fl. oz. Imp., into f5, U. S. Solution.— 1 fl. oz. = .96 f^. And 6 X .96 = 5.78. Therefore 6 fl. 02. = 5.76 f 5. 42 Pharmaceutical Arithmetic But since our customary measures are not graduated in dec- imal fractions of each unit, convenience in measuring requirea that such decimal fractions be reduced, so as to express the amount in integers of a lower unit (aenomination). Hence .76 fj should be reduced to f3; and should there be a fractional f3 in the product, this fractional fluid drachm should be reduced to minims, as follows: .76 X 8 = 6.08. Hence .76 f5, = 6 f 3 and .08 f3j and since .08 X 60 = 4.80, .08 f3 = 48 Tli. Then 5.76 f 5 = 5 f 5 6 f 3 4.8 in.. 29. Convert 6 fj into Imp. measure, Ans 6 fl. oz. 2 fl. drm. 30. Convert 1 Imperial pint into pints, f3, f3 and ni, of U. S. measure. Ans. 1 O. 3 f5 1 f3 36 in.. (U. S.) 31. Convert 3 Imperial pints into Apothecaries' measure. 32. Convert 3 pints U. S. Into Imperial measure. 33. Convert 6 pints 6 fluid ounces 6 fluid drachms Imp., into U. S. apothecaries' measure. Suggestion. — Reduce to fluid drachms before converting into U. S. measure. And should the final answer contain a fraction of a drachm, reduce this to minims. 34. Convert the Imperial measures In the following Canad- ian prescription into U. S. apoth. measure: Liq. ezt. bellad. 10 fl. oz. Camphor 1 oz. Dist. water 2 fl. oz Alcohol q. s. ad 20 fl. oz. Note. — If the Ingredients were all liquids, it would be allow- able to read f3 in place of fl. oz. A somewhat larger volume of liniment would be dispensed in that case, but the relative pro- portions of the Ingredients would not be changed. In the pres- ence of solids, however, it is different; for the relationship be- tween the Imperial ounce and the Imperial fluid ounce on the one hand, and the apoth. ounce and the apoth. fluid ounce on the other, is not the same. See page 30 and page 62. However, the difference Is so small that it is usually ignored. 4. Metric and ImperiaIv Measures. The following eqmvalents may be used: 1 Imp. fl. oz. = 28.39 mils or c.c. 1 c.c. = 16.9 Tfl (Imp.) But the equivalents are taldom required and need not be memorized. Should it be necessary to convert metric measure into Imperial, or vice versa, and the equivalents are not at hand, convert the measure given to apothecaries' and this to the meas- ure required, as indicated under problems No. 35 and No. 36. Relationship of Systems 43 35. Convert 8 fl. oz. (Imp.) Into c. c. Solution.— 8 X .96 = 7.68. .'. 8 fl. oz. = 7.68 fj. i7.68 X 29.57 = 227. .'. 7.68 fS = 227. c. C. 36. Convert 500 c. c. into fl. oz. Process.— 500 -r- 29.57 = number of fj. Number of f5 -r- .96 = number of fl, oz. 37. Convert 40 Tit (ImP-) ii^to c. c. Ans. 2. 3 c c. 38. Convert 1 L. Into Imperial measure — pints, fluid oimces, €ta Ana 1 pint 15 fl. oz. 1 fl. drm. 45 ITI,. 39. Convert 50 gallons (Imp.) Into Liters. Note. — ^The sign . * . stands for then, hence or therefore^ Approximate Measures. The following domestic measures are in common use: I Tumblerful = 8 f ^ = nearly 240 c.c. I Teacupful = 4 f § = nearly 120 c.c. I Wineglassful = 2 f 5 = nearly 60 c.c I Tablespoonful = i/^ to % f 3 = 15 to 20 c.c. I Dessertspoonful = ^4 to J4 f 5 = 7-5 to 10 cc I Teaspoonful z= ^ to J^ f 5 = 3-75 to 5 c.c. I Drop (watery liquids) = i V[\, = about 0.06 c.c. I Drop (alcoholic liquids) := j/2 VI = about 0.03 cc. But the U. S. P. IX gives the following as sufficiently accurate: i tablespoonful =^ fl. ounce or 15 mils; i des- sertspoonful = 2 fluidrachms or 8 mils ; i teaspoonful == i fluidrachm or 4 mils. The first equivalent given in the table is the old, and the second, the new, and in conformity with the actual ca- pacity of spoons. In case of very active medicines this dis- crepancy between the equivalents commonly used and the actual capacity of spoons used by the patient, should not be 44 Pharmaceutical Arithmetic overlooked. If it is intended that a I f 3 dose be admin- istered, the directions should read ''^ teaspoonful, etc." Another source of error in dosage is the use of a dessert spoon in place of a tablespoon. It should be remembered also that the drop is not a definite volume, and that it seldom measures exactly i "HI, even in case of pure water. The size of the drop is depend- ent not only on the nature of thel iquid, but is influenced by the temperature, and by the size and shape of the mouth of the vessel from which the drop is emitted. In case of medicine droppers, which are much used in prescription work, the drop varies directly with the outside diameter of the lower extremity of the tube. Problems. 40. Write a prescription for 32 teaspoonful doses, each to contain i^ TTt of tincture of aconite, and enough compound elixir of taraxacum to make the proper volume. 41. B Antim. et potass, tart. gr. iv Liq. amm, acet. fi iv Spt. aeth. nit. fSij Tr. aconiti f3 j Syr. q. a. ad f5 viij Teaspoonful 3 times a day. Calculate the single dose for the first four ingredients. 42. 3 Tincturae veratri viridis 3 Spiritus aetheris nitrosi 30 Liquoris potassii citratis 20 Syrupi zingiberig q. s. ad 240 Calculate the amount of each of the first three ingred- ients in a tablespoonful dose. Relationship of Systems 45 43. Write a metric prescription for 8 wineglassful doses, each to contain 0.5 c.c. of tincture nux vomica, and enough compound infusion of gentian (Br.) to make the proper volume. 44. 5 Ac. hydrocyan. dil. f3 ij Aq. amygd. amar. q. s. ad fS iv Teaspoonful. (a.) How much dilute hydrocyanic acid in each f3 dose? (b.) How much of the acid would be administered for a dose if a common teaspoon were used? Metric and Apothecaries' Weights. OfSdal Equivalents. Approximate Equivalents. 1 gr. = 64.8* mg. 1 gr. =65 mg., or .065 Gm, 1 § = 31.1035 Gm. 1 § = 31 Gm. 1 Gm. = 15.4324 gr. 1 Gm. = 15.4 gr. In reducing gr. to Gm., use the equivalent, i gr. = .065 Gm., rather than the equivalent, i Gm. := 15.4 gr. For multiplication involves less work than does the process of division. In computing doses, the equivalent, I gr. = 65 mg., is sufficiently accurate, and is generally used. Problems. Use Official Bquivalents. Convert 6§23i 9 12 gr. into Gm. Ans. 196.47 Gm. 45. (a.) Convert fV gr. into mg. ; (b.) 6.4 gr. into mg. Ans. (a.) 12.1 mg.; (b.) 414.7 n.g. 'The official table gives 64.79; but 88 the accurate equivalent is 64.7089, 64.3 is more nearly correct than 64.79. tAccurately, 15.43235639 gr. 46 Pharmaceutical Arithmetic 46. (a.) Convert 64 mg. into gr. ; (b.) 6.042 Gm. into gr. 47. Convert 200 Gm. into apothecaries' units. Ans. 6 5 3 3 I 9 63^ gr. Use Approximate Equivalents. 48. Give the following doses in metric weight: Arsenous acid. ^ to ^ gr. ; atropine, ^^ ^<^^ gr- ; ext. bel' ladonna leaves, ^ to 3^ gr. ; ext. rhubarb, i to 10 g^. ; ferric dtrate 3 to 20 gr. ; salol, 5 to 30 gr. ; sodium salicylate, 5 to 60 gr.; cascara sagrada, 15 to 60 gr. ; coca, ^ to 2 3; Rochelle salt, i to 8 3. Ans. 49L 9 Bismuthi subcarb. 3 ii j = 11.7 Gm. Morphinae sulph. gr. j = 0.06S Gm. Pulv. aromat. 3j = 3.9 Gm. Make 12 powders. (a.) Convert to metric weight, (b.) How many mg: of morphine sulphate in each dose? 50. Express the following doses in apothecaries' weight : Aconitine, .0001 Gm. ; hyoscyamine sulphate, .0008 Gm.; ipecac (as emetic), 1.4 Gm., jalap, .65 to 1.3 Gm, ; ext. nux vomica, .008 to .016 Gm. ; antimonial powders, .2 to .52 Gm. ; quinine sulphate, .065 to 1.5 Gm. ; effervescent mag- nesium sulphate, y.77 to 31.1 Gm. 51. Express the quantities in the following prescrip- tion in apothecaries' weight : 9 Qnininae sulphatis Strychninae sulphatis Ferri phosphatis Make 50 pills. Ans. 25 = 2 3 10 gr, 06S = lgr. SO =-=1323 Relationship of Systems 47 7. Metric and Avoirdupois Weights. Official Equivalents. Approximate Equivalents. 1 gr. = 64.8* mg. 1 gr. — 65 mg. 1 oz. = 28.35 Gm. 1 oz. = 28y2 Gm. 1 lb. = 453.592 Gm.t lib. =:454Gm. 1 Gm. = 15.4324 gr. 1 Gm. = 15.4 gr. 1 Kg. = 2.2 lb.$ 1 Kg. = 2.2 lb. The metric pound is 500 Gm., or ^ Kg. Problems. Use Official Equivalents, 52. Convert (a.) 5 oz. into Gm. ; (b.) 3 lb. into Gm. ; (c.) 5 lb. into Kg. ; (d.) 5 lb. 5 oz. 200 gj. into Gm. Ans. (d.) 2422.66 Gm. 53. Convert (a.) 60 Gm. into gr. ; (b.) 600 Gm. into oz. ; (c.) 6 metric pounds into av. pounds ; (d.) 5675.45 Gm. to avoir, weight. Ans. (d.) 12 lb. 8 oz. 85.6 gr. Use Approximate Equivalents. 54. Express the quantities in the following formula foe menthol plaster in metric weight : Menthol 2 oz. Yellow Wax 2 oz. Resin 1 oz. Lead Plaster 15 oz. 55. (a.) Convert 5 lb. into Gm. ; (b.) 5 oz. into Gm.; (c.) 5 gr. into mg.; (d.) 50 Gm. into gr. 56. Convert the quantities in the formula for com- pound cathartic pills into avoirdupois weight: Com. ext. colocjmth 800 Gm, Mild mercurous chloride 600 Gm. Resin jalap 200 Gm. Gamboge 150 Gm. •Accurate equivalent la 64.7989 mg. t Accurately, 453.592653 Gm. ^Accurately, 2.20462 lb. 48 Pharmaceutical Arithmetic ■57. A druggist buys the following list of goods: 5 Kg. Powd. acacia. Yz Kg. Iodoform. 2 Kg. Ether. 100 Gm. Lithium citrate. 25 Gm. Amyl acetate. 500 mg. Aconitine. (a.) Convert these weights into avoirdupois units, (b.) Box and packing being figured at 8 Kg., how much would the gross weight be in avoir, units ? Ans. 34 lb. 2 oz. Avoirdupois (or Imperiai,) and Apothecaries' Weights. Equivalent : I gr. apoth. = I gr. av. or Imp. Note. — Since the grain apoth. is identical with the grain avoir., any weight in one of these systems can be converted into weight in the other, by reducing to grains, and then to the higher units of that other system. Thus, 6 5 = 2880 gr. = 6 oz. 255 gr. 6 oz. = 2625 gr. = 55 3 3 2 3 5 gr. To use the equivalent, 1 5 = 1.1 oz.* would be practically no shorter (since the fraction would have to be reduced to smaller units), and the answer would be only approximately correct. Problems. 58. (a.) Convert 2 § into avoir, weight. (b.) Convert I4}i § into avoir, weight. 59. (a.) Convert 1533329 10 gr. into avoir, weight, (b.) Convert 35 30 2 9 15 gr. into Imp. weight. Ans. 60. U Ext. Stramonii Slj = 120 gr. . Adipis Benz. Jij = 2 oz. 85 gr. Express quantities in avoir, weight. •13 = 1.1 07» av or Imp. (accurately, 1.097143). Relationship of Systems 49 Note. — Ointments, dusting powders, etc., are usually prescribed in quantities of 1 5 or more. But the 2 3 weight is the largest in the common set of prescription weights, and in many stores larger apoth. weights are not at hand. The proper procedure then is to reduce the apoth. weight to avoir, weight, in order that the coun- ter weights, which are avoir., may be utilized. For instance, 2 5 of petrolatum is to be weighed. 2 5 = 2 oz. and 85 gr. So the 2 oz. weight of the counter set is used, together with a 1 3 weight, a 1 3 weight, and a 5 gr. weight of the prescription set. '/o weigh out 1 oz. when 1 5 is prescribed, involves a change in proportion of active medicine to base or diluent — a change which the pharmacist is not justified in making, except with the consent of the physician. 61. n Sodium sulphate 5 iij Sodium chloride 5 viij Linseed, powd. S viij Suppose av. oz. were weighed in place of the apoth. ounces. How much difference would it make in the weight of the mixture? Note. — ^In this prescripion the 5 is the only unit used. And should avoir, ounces be dispensed in place of apoth. ounces, the proportions would not thereby be changed. There would still be 3 parts of sodium sulphate to 8 parts of the other ingredients. So there could be no serious objection to such a change. 62. (a.) Convert 3 oz. av. into apoth. weight Ans. 25 5 3 2 9 12I/2 gr- (b.) Convert 5 oz. Imp. into apoth. weight. Ans. 4 § 4 3 I 9 yy2 gr. (c.) Convert 3 oz. 300 gr. into apoth. weight, (d.) Convert 7 \\ oz. into apoth. weight. 63. A pharmacist dispenses 3 iv of aristol from a I oi package. How many grains remain? How many 3? 64. How many i9- doses of bismuth sub-nitrate could be dispensed from a 4 oz. package of the salt ? CHAPTER III. VOLUMES AND WEIGHTS. A. Specific Gravity (Relative Density, Specific Weight) By weighing equal volumes of various substances the fact is made apparent that hardly two substances can be found which weigh exactly alike volume for volume. Gly- cerin weighs one and one-fourth as much, volume for vol- ume, as water. Ether is not quite three-fourths as heavy as water. Mercury is nearly twenty-three times as heavy as lithium, but not quite twice as heavy as iron. Thus the relation between weight and volume — in other words, the density, — differs with nearly every substance, but is always (under like conditions) the same for the same substance. This being true, density serves as a characteristic, by which substances may be identified, or their purity be determined. The most convenient way of expressing the density of a substance is by comparison with the density of another substance. The standard selected for such comparisons, for solids and liquids, is pure water*, the density of which is taken as unity — as i.ooo. The density of any substance twice as heavy as water may then be expressed by 2.000 ; the density of a substance one-half as heavy as water, by .5 ; — ^these numbers indicating not weights, but ratios between weights. When density is thus expressed (by comparison), it is properly called relative density; but its synonyrp, specific gravity, is more generally used, and will be used in the suc- ceeding pages. To repeat : Density is the relation which the weight of a body bears to its volume. Specific gravity is the density of a body as compared with the density of water, or with some other arbitrarily established standard. Since the volume, and hence the density, of a body varies with the temperature, it is necessary to have a stand- *For gases, bjdrogen at 0°C ia taken as the standard of density. Specific Gravity $1 ard temperature for specific gravity determinations. The temperature adopted by the U. S. Pharmacopoeia is 25° C, both for the standard — water — and for the substance of which the specific gravity is to be determined. This is ex- pressed by 25" C. 25" C In Europe it is customary to make the determination at [or near] 15° C, but to compare with water at 4" C, the temperature at which water reaches its g^reatest density. These temperatures are expressed by 15° c. 4°C. Calculating Specific Gravity, To find the specific gravity of a body divide its weight by the weight of an equal bulk of water. Fractions in the quotient are stated as decimals ; and in case these are inter- minate, they are usually carried to the third place. The method employed in the determination of the equal bulk of water, varies with the nature of the substance under examination. In case of liquids, a small bottle is com- pletely filled with water at the proper temperature, and weighed. Then the same bottle is filled with the liquid, and weighed. In each case the weight of the bottle itself is subtracted, giving in the latter case the net weight of the liquid, and in the first case the net weight of an equal bulk of water* Problem. — A bottle holds 96.54 Gm. of water, and 94.78 Gm. of a certain oil. What is the sp. gr. of the oil? Operation: 94.78 -~ 96,51 = .981 (wt. of body of oil) -r- (wt. equal bulk of water) = sp. gr. The specific gravity of a solid, in mass, is found as fol- lows: The body is weighed in air, in the usual manner. Then it is weighed in water being suspended in the latter from a balance arm by means of a hair. In water the body •Another method is to make use of floats or buoys which indicate the sp. fX- of the liquid by the depth to which they sink. Such instruments are called hydrometers. The graduated scale on a modem hydrometer indicates the sp. gr. directly; but the older instruments carry arbitrary scales, some of which, notably those of Beaume, are still in use. For relation of degrees Beaume to sp. gr. see chapter 21, 52 Pharmaceutical Arithmetic is found to weigh less than in air. The difference between the two weights (the loss) is due to the buoyant force of the water, and this is exactly equal to the weight of just as much water as the body displaces. In other words, the loss is exactly equal to the weight of an equal bulk of water. Problem. — A piece of ore weighs 5.4 Gm. in air. Suspended in water, from the arm of a balance, it weighs 4.01 Gm. Calculate the sp. gr. of the ore. Operation. — ^Wt. in air = 5.4 Gm. Wt. in water = 4.01 Gm. Loss in water = 1.39 Gm. (wt. of equal bulk of water). Then 5.4 Gm. -f- 1.39 [Gm.] = 3.884 (sp. gr.) If the solid is found to be lighter than water, a sinker must be used to make complete immersion possible, thus making the calculation more complicated.* If the solid is soluble in water, another liquid in which the solid is in- soluble must be selected, and this taken cognizance of in the calculation.t Problems. I. A certain bottle holds 500 Gm. of water. Of a cer- tain solution it holds 650 Gm. Calculate the sp. gr. of the solution. Ans. 1.3. •Problem. — A piece of paraffin weighs 2.895 Gm. Attached to sinker the total weight ia 14.086 Gm. The sinker and paraffin in water weigh 8.258 Gm., while the sinker by itseU in water weighs 8.652 Gm. Calculate sp. gr. of the paraffin. Operation.— Vft. of sinker in air 11.191 Gm. Wt. of sinker in water 8.652 Gm. Wt. of water displaced by sinker 2.539 Gm. Wt. of paraffin plus sinker in air 14.086 Gm. Wt. of paraffin plus sinker in water 8.258 Gm. Wt. of water displaced by paraffin plus sinker 6.828 Gm. Wt. of water displaced by sinker 2.539 Gm. Wt. of water displaced by paraffin 3.288 Gm. Then 2.895 (wt. of paraffin in air) ■— 3.289 (wt. of equal bulk of water) = .880 (sp. gr. of paraffin.) f Problem. — A crystal of alum weighs .1.824 Gm. in air, and 1.901 Gm., sus- pended in oil of turpentine, having a sp. gr. of .860. What is the sp. gr. of the alum? Operation.— Wt. of alum in air 3.824 Gm. Wt. of alum to oil turpentine. — 1.901 Gm. Wt. of oil turp. displaced. 1.923 Gm. Hien 3.824 Gm. -J- 1.923 [Gm.] = 1.988 (sp. gr. of alum, if oil turpentine were standard of density). Finally, 1.9SS X .860 (sp. gr. of oil turp.) = 1.700 (sp. gr. of alua, referred t« irater, the standard of density). specific Gravity 53 2. A certain piece of metal weighs lo.io Gm. in air, and 8.65 Gm. suspended in water. What is the sp. gr. o£ the metal? Ans. 6.965. 3. A certain iron cylinder holds 50 lb. of mercury. Of water it holds 3.69 lb. What is the sp. gr. of mercury? 4. What is the sp. gr. of a crystal which weighs 2.34 Gm. in air, and 1.18 Gm. suspended in water? 5. A certain bottle holds 245 gr. of water. Of oil of peppermint it holds 220.5 gr. What is the sp. gr. of the oil ? 6. A certain bottle holds 3 oz. (av.) of water. Of nit- ric acid the same bottle holds 120.2607 Gm. What is the sp. gr. of the nitric acid. Ans. 1.4. Note. — When a denominate number is to be divided by another denominate number, both must be in the same unit (denomination.) 7. An av. oz. of a volatile oil measures 35 c.c. What is the sp. gr. of the oil ? Solution. — 1 av. oz. = 28.35 Gm.; and 35 c.c. of water weigh 35 Gm. Hence— 28.35 Gm. -^ 35 = .81 (sp. gr.). 8. The weight of 5 c.c. of a certain liquid is lOO gr. What is the sp. gj? Ans. 1.296. 9. If a minim of water weighs .95 gr., and an av. oz. of an oil measures 500 TTL, what is the sp. gr. of the oil? Solution.— 500 n\, of water weigh 500 X .95 = 475 gr. 500 TTl, of oil weigh 1 oz. = 437.5 gr. Then 437.5 gr. -r- 475 gr. = .921 (sp. gr.) 10. What is the sp. gr. of a liquid i lb. of which meas- tires ^ L? Table of Specific Gravities. The approximate specific gravity of the more common liquids should be memorized. Water 1.00 Alcohol 0.81 Glycerin 1.25 Chloroform 1.5 Ether 0.7 Ammonia Water 0.96 Sulphuric Acid 1.83 Hydrochloric Acid 1.16 Nitric Acid 1.4 Syrup 1.3 Mercury 13.5 Etc. 54 Pharmaceutical Arithmetic Specific Volume. Specific volume expresses the same fact expressed by specific gravity ; namely, the density of a body as compared with the density of water. To find the specific gravity, the weight of a body is compared with the weight of an equal volume of water. To find the specific volume, the volimie of a body is compared with the volume of an equal weight of water. Specific vol- ume is, therefore, the reciprocal of specific gravity. The greater the specific gravity of a body, the smaller is its spe- cific volume; and vice versa. Calculating Specific Volume. — Divide the volume of the body by the volume of an equal weight of water. vol. of body Sp.V.= vol. of equal wt. of water. Fractions In the quotient are expressed in decimals, zixid are carried to the third place, as a rule. 11. A certain weight of oil measures 60 c.c. The same V, eight of water measures 52 c.c. What is the specific vol- ume of the oil ? 60 C.C. -f- 52 c.c. = 1. 153 Vol. of oil. -V- Vol. of equal wt. of water = Sp. V. 12. A certain weight of a certain acid measures 80 c.c. The same weight of water measures 4 f J. What is the sp. V. of the acid ? Ans. .676. Calculating Specific Volume from Specific Gravity. Rule. — ^Divide the sp. g^. into i.ooo. The quotient is the sp. v. 1.000 Sp. V.= Sp. Gr. 13. Glycerin has a sp. gr. of 1.25. What is its sp. v. ? i.oo^- 1.25 (sp. gr.) = .800 (sp. V.) 14. Sulphuric acid has a sp. gr. of 1.8. What is its sp. T.? Volumes to Weights 55 15. Ether has a sp. gr. of .716. What is its sp. v.? Ans. 1.396. Calculating the Specific Gravity from the Specific Volume. Rule. — Divide the sp. v. into i.ooo. The quotient is the sp. gr. 16. Chloroform has a sp. v. of .677. What is the sp. gr ? I.oo -V- .677 (sp. V.) = 1.476 (sp. gr.) B. Reducing Volume to Weight. I. Mils or Cubic Centimeters to Grammes. The weight of i c.c. of water is i Gm. The specific gravity of water is i.ooo, it being the standard of density. One c.c. of any liquid having a specific gravity of i.ooo, will, therefore, weigh i Gm. One c.c. of a liquid having a sp. gr. of 2.000, will weigh 2 Gm. ; and one c.c. of a liquid having a sp. gr. of .75, will weigh .75 Gm. In short, the specific gravity of a liquid indicates the weight in Grammes of one cubic centimeter of it. Then any volume expressed in cubic centimeters may be converted into weight expressed in Grammes by multiplying the number of cubic centimeters by the specific gravity : — No. of c.c. X sp. .gr. = No. of Gm. Problems. 17. What is the weight in Gm. of 500 c.c. of glycerin, having a sp. gr. of 1.25? 500 X 1.25 = 625. (No. of C.C.) X (sp. gr.) = (No. of Gm.) 56 Pharmaceutical Arithmetic 18. 2 L. of sulphuric acid, sp. gr., 1.826, weigh how many Gm.? Ans. 3652 Gm. 19. What is the weight in Gm. of 650 c.c. of alcohol, sp. gr., .82 ? Ans. 533 Gm. 20. Ether has a sp. gr. of .716. How many Gm. will 5.67 c.c. weigh? Ans. 4.0597 Gm. 21. Calculate the weight in Gm. of 25 L. of syrup, sp. gr., 1.3. Ans. 32500 Gm. 2. Fluid Ounces to Grains. The weight of one fl. ounce of water, at 25" C., is 454.6 gr. Water being the standard of density, it follows that any liquid having a sp. gr. of i.ooo, will weigh 454.6 gr. per fl. ounce ; that a liquid with a sp. gr. of 2.000, will weigh twice 454.6 gr. per fl. ounce ; and that a liquid with a sp. gr. of .900, will weigh .9 of 454.6 gr. per fl. ounce. In every case 454.6 multiplied by the sp. gr. of the liquid will give the weight in grains of one fl. ounce. Then any volume in fl. ounces may be reduced to grains by the following rule : 454.6 gr. X sp. gr. x no. of f § = wt. in gr. 22. Calculate the weight in gr. of 2 pints 3 fl. ounces of nitric acid, sp. gr., 1.4. Solution. — 454.6 gr. X 1.4 = 636.44 g^. (wt. in gr. of I fl. ounce.) 2 pints 3 fl. ounces = 35 fl. ounces. Then, 636.44 gr. X 35 = 22275 gr. Ans. Usually the weight in gr. is subsequently to be reduced to weight in higher units. Note — ^When decimals in a product are dropped, this should be done with the least error: — 1.158 for instance, should be given as 1.16, not as 1.15, 2^. (a.) Find weight in gr. of 14 f ^ of mercury, sp. gr., 13.5. (b.) Express in higher units, avoirdupois. 24. (a.) Calculate weight in av. lb. of 36 gallons of alcohol, sp. gr., .81. (b.) Calculate weight in av. lb. of i pint of syrup, sp. gr., 1.31. 25. Calculate weight, in av. lb. of i pint of water, Ans. I lb. 273.6 gr. Volumes to Weights 57 26. Calculate weight in av. lb. of i pint of benzine, sp. gr., .67. Ans. .696 lb. 2y. Calculate weight in av. lb. of i pint of sulphuric acid, sp. gr., 1.83. Ans. 1.9 lb. [Is it a fact that "a. pint is a pound"?] 28. Calculate weight, in units of the apoth. system, of I O. 3 f 5 2 f 3 40 TTL of alcohol, sp. gr., .81. 29. Calculate weight in Gm. of i gallon of ammonia water, sp. gr., .96. Ans. 3632.6 Gm. 3Q. Calculate weight in av. lb. of 8 L. of stronger am- monia water, sp. gr., .9. 3. Minims to Grains. Since I fl. ounce equals 480 minims, and since I fl. ounce of water weighs 454.6 grains, each minim of water weighs 454-6. grains. 480 454.6 But is too cumbersome a fraction for speedy work ; 480 a decimal fraction would be more convenient. 454-6 = .947. 480 And this fraction may, for most practical purposes, be rounded off to .95. Then, if .95 gr. is taken as the weight of I ITt of water, the weight of i ttl of any liquid must be .95 gr. X sp. gr. ; and the rule for reducing ttl to gr. must be — •95 gr. X sp. gr. X no. of TT), = weight in gr. Problems. 31. I'ind weight in grains of 160 TH, of solution of fer- ric chloride, sp. gr., 1.3. 58 Pharmaceutical Arithmetic Solution.—' .95 gr. X 1.8 = 1.235 gr. Wt of 1 TH, of -water X sp. gr. r= wt. of Im, of solution. Then. 1.235 gr. X 160 = 197.6 gr. (Ans.) 32. Find weight in gr. of 75 Tti of sulphuric add, sp. gr., 1.8. Ans. 128.25 gr. 33. Find weight in gr. of 200 Til of alcohol, sp. gr., .81. 34. Find weight in gr. of 120 in. of chloroform, sp. gr., 1.48. Ans, 168.7 gr. 4. Other Units Apoth. Measure to Apoth. Weight. One of the chief advantages of the apothecaries' systems lies in this, that the three units of volume most commonly used in prescription work, and the three most commonly used weight units, are parallels. Thus, the smallest unit of volume, the minim, has its parallel in the grain, which, as has been seen, is, however, not exactly equivalent to a minim ; and just as 60 minims make i fl. drachm, so 60 grains make I drachm ; and as 8 fl. drachms make i fl. ounce, so 8 drachms make i ounce. It follows, then, that if i minim of water weighs .95 grains, i fl. drachm of water must weigh .95 drachms (3), and i fl. ounce, .95 ounces (§). Hence the following rule to reduce fl. drachms of a liquid to weight in drachms : .95 3* X sp. gr. X No. of f 3 = weight in 3. And the following rule to reduce fl. ounces to ounces : •95 §* X sp. gr. xNo. of f 5 = weight in. §. Note. — If the student cannot harmonize this rule with the one given on page 57, he should carefully review pages 55, 56 and 57. Problems. 35. What would be the weight in ^ of I O. 6 f^ of stronger ammonia water, having a sp. gr. of .9? Ans. 18.8 5 26. What would be the weight in 3 of 3^^ f 3 of lactic acid, having a sp. gr. of 1.2? Ans. Nearly 4 3. 'Exactly, .947 gr., as «zplairMl on i>age SI, Weights to Volumes 59 ^y. What would be the weight in 3 of 7^ f 3 of phos- phoric acid, sp. gr., 1.71 ? 38. What would be the weight in § of 2 O. of diluted alcohol, sp. gr., .93? Reducing Weight to Volume. I. Grammes to Cubic Centimeters or Mils. As was explained on page 55, the specific gravity expresses the weight in Grammes of one cubic centimeter. Thus the expression, glycerin, sp. gr., 1.25, means that each cubic centimeter of glycerin weighs 1.25 Gm. Then 1.25 Cm. of glycerin will measure i cubic centimeter ; 2.5 Gm. of glycerin will measure 2 cubic centimeters; and 50 Gm. of glycerin will measure as many cubic centimeters as 1.25 Gm., the weight of i cubic centimeter, is contained in 50 Gm. Hence the rule — Wt. in Gm. -r- sp. gr. = vol. in c.c. (mils). Problems. 39. A solution of ferric sulphate has a sp. gr. of 1.32. Calculate volume of 358.5 Gm. Ans. 271.59 c.c. 40. A formula calls for 246 Gm. of nitric acid, sp. gr., 1.4. But on account of the corrosive action of the acid on the balances, it is much more convenient to measure the acid than to weigh it. How many c.c. should be used? 41. A formula calls for 870 Gm. of sulphuric acid, sp. gr., 1.826. How many c.c. should be used? Ans. 476 c.c. 42. In making ferric chloride, 300 Gm. of hydrochloric acid is to be used. The acid having a sp. gr. of 1.16, how many c.c. should be measured out? 43. A certain solution having a sp. gr. of 1.4, is directed to be given in 2 Gm. doses. What would be the dose in mils ? [In Europe it is customary to compound all medicines, liquids as well as solids, by weight. It is also customary to state the dose in weight units.] 6o Pharmaceutical Arithmetic 44. A certain tincture is directed to be given in 3 Gnu doses. It has a sp. gr. of .86. What should be the dose in TlX? 45. In a certain formula i o oi nitric acid, sp. gr., 1.403, is required. How many c.c. should be used? Suggestion, — Reduce I o to Gm. ; then to c.c. 46. A certain formula calls for i Kg. of phosphoric acid, sp. gr., 1.71. How many c.c. should be used? 47. A formula calls for 5 oz. of nitric acid, sp. gr^ 14. How many c.c. should be used? 2. Grains to Fluid Ounces. As has been shown on page 56, the weight of i fi. ounce of any liquid may be calculated if the specific gravity is known,, this being used as a multiplier for 454.6 gr., the weight of I fl. ounce of water. Then, if the weight of i fl. ounce of the liquid is known, any given weight can be reduced to fl. ounces by dividing the given weight by the weight of i fl. ounce. This may be expressed as follows : Wt. of liquid in gr. -^ (454.6 gr. x sp. gr.) = voL infS. [And, since multiplying the divisor is the same as dividing the dividend, the no. of f S may be found also by dividing the weight of the liquid in gr. first by 454.6 gr., then by the sp, gr. Wt of liquid in gr. -r- 454.6 gr. -h sp. gr. = vol. in fj. Problems. 48. Calculate the volume in f 5 of I lb. of commercial ether, sp. gr., .725. Solution. — 1 lb. = 7000 gr. 454.6 gr. X .725 = 328.6 gr. wt. 15 of water X sp. gr. = wt. of 1 f5 of ether. Then, 7000 gr. -i- 328.6 (gr.) = 21.3. (Ans.) (wt. given ) -r- ( wt. of 1 f 5) ^ no. of f J. 49. Find volume in f 5 of i lb. of water. Weights to Volumes 6x 50. Find volume in f 5 of I lb. of chloroform, sp. gr., 1.48. Ans. 10.4 f §. 51. Find volume in f 5 of I lb. of mercury, sp. g^., 52. Find volume in f o of i lb. of oil of lemon, sp. gr., .858. Ans. 17.9 f 5, or 17 f 5 7 f 5 12 Til. 53. Find volume in f § of 5o 3^ i 9 10 gr. of solution of ferric chloride, sp. gr., 1.3. 54. Find volume in pints of 5 lb. of nitric acid, sp. gr., 1.4. 55. Find volume in c.c. of 4 lb. of hydrochloric acid, sp. gr., I. 158. Ans. 1568 c.c. 56. Find volume in L. of 50 lb. of glycerin, sp. gr., 1.25. 57. Find volume in pints of 25 Kg. of glycerin, sp. gr., 1.25. Ans. 42.26 pints. 58. One pound (av.) of ether, sp. gr., .716 would fill how many 2 f § bottles? Ans. Nearly 11. 59. One hundred pounds of mercury, sp. gr., 13.53 would fill how many pint bottles? Ans. 7. 11 pints. . 60. How many fl. ounces in 2 Kg. of chloroform, sp. gr., 1.48? Ans. About 45. 61. A carboy of stronger ammonia water, sp. gr., .9, contains 115 lb. of the water. How many pints does it con- tain ? How many Liters ? 62. A carboy of sulphuric acid, sp. gr., 1.8 contains 142 Kg. How many cubic centimeters does it contain? How many fl. ounces? How many Liters? How many gallons ? 63. A druggist buys glycerin, sp. gr., 1.25, at 45c. per lb. and sells a pint at 80c. What is his profit on each pint ? 64. A druggist buys chloroform, sp. gr., 1.48 at 70c. per lb., and sells a Liter for $2.00. Does he gain or lose? How much? Ans. He loses 28.2c. 3. Grains to Minims. Since I Til, of water weighs .95 gr., .95 gr. multiplied Ey the sp. gr. gives the weight in gr. of i iTt of any liquid and—* No. of gr. given -f- (.95 gr. x sp. gr.) = vol. in HI,.* •Or, No. of gr. given -f- .96 gr. -j- sp. gr. = vol. in Vl^, 62 Pharmaceutical Arithmetic 65. Find volume in minims of 200 gr. of c.p. chloro- form, sp. gr., 1.49. Solution.— .95 gr. X 1.49 = 1.4155 gr. (Wt. of 1 in. o* water) X (sp. gr.) = (wt. of 1 ITL of chlorofonn.) Then— 200 gr. -^ 1.4155 gr. = 141. (Ana.) No. of gr. given -r- wt of 1 TH, = No. of Ttl. 66. Find volume in minims of 90 gr. of lactic acid, sp. gr., 1.2. Ans. 79 TTl. 4. Drachms and Ounces to Fluid Drachms and Fluid Ounces. As was explained on page 58, i f 3 of water weighs about .95 3 ; and i f 5 of water, about .95 ^. Then — ,No. of 3 given -i- (.95 3 x sp. gr.) = vol. in f 3 ; and No. of 5 given -r- (.95 3 x sp. gr.) = vol. in f 5- The answers will be slightly high, but sufficiently accur- ate for practical purposes. The student should explain why these answers are slightly higher than answers obtained by rules on page 60. 6y. Find volume in f 3 of 6 3 of bromine sp. gr., 2.99. 68. Find volume in f 5 of 8 § of phosphoric acid, sp. gr., 1.71. Ans. 4.9 f 5, or 4 f 5 7 f 3 21 Til. 69. Find volume in c.c. of 8 5 of phosphoric acid, sp. gr., 1.71. Ans. 147.7 c.c. 70. Find volume in f § of 250 Gm. of phosphoric add, sp. gr., 1. 71. 5. Imperial Ounces to Fluid Ounces. An Imperial or avoir, oz. of water measures an Imperial fluid ounce. Hence — No. of Imp. oz. -r- sp. gr. = Imp. fl. oz. 71. Find volume in Imp. fl. oz. of 10 Imp. oz. of gly- cerin, sp. gr., 1.25. Ans. 8 Imp. fl. oz. CHAPTER IV. REDUCING AND ENLARGING FORMULAS. Let us suppose that just 25 Gm. of compound morphine powder is wanted. The formula to be used is that of the U. S. P., 1890, as follows: Morphine sulphate I Cm. Camphor 19 Gm. Glycyrrhiza 20 Gm. Precip. calcium carbonate.. .20 Gm. To make 60 Gm. Now 25 Gm. is f ^ of 60 Gm. So we must multiply the quantity for each ingredient by f ^ , in order that the mixture may weigh 25 Gm. in place of 60 Gm. That is, we must divide by 60, which would give us the amount to be used for I Gm. of powder, and then multiply by 25, because we wish to make 25 times i Gm,, namely, 25 Gm.* From the above reasoning we may deduce the following general rule: To reduce or enlarge a formula, multiply the quantity of each ingredient by a fraction of which the quantity [of the preparation] to be made is the numerator, and the quantity the formula makes is the denominator. In practice this fraction is always reduced to the simplest expression ; f-^ be- comes i ; ^ becomes -i^-; -/JV^ becomes ^V or .6. Problems. I. How much vinegar of squill, and how much sugar are required to make 600 mils of syrup of squill? Formula : Vinegar squill 450 mils Sugar 800 Gm. Water q. s. ad 1000 mils * In practice cumbersome fractions are to be avoided when possible. The practical druggist would make 30 Gm. In place of 25 ; and since 18 = %, would avoid the lengthy calculation, and save enough time to compensate for the extra 5 Gm. of powder, should the latter be wasted. 64 Pharmaceutical Arithmetic The formula is for looo c.c, but we wish to malce 6oa C.C., which is yWV or .6 of lOoo c.c. Hence each quantity must be multiplied by t^ = *'^» Operation: 450 C.C X .6 = 270 c.c. 800 Gm. X .6 = 480 Gm. 1000 cc. X .6 = 600 c.c. 2. How much iodine is required to make I pint of a tincture, a Liter of which contains 70 Gm. of the constitu- ent? Solution.— 1 0. = 16 f 5; and 1 f 5 = 29.57 c. c Accordingly 1 O. = (16 X 29.57) = 473 cc. Then 70 Gm. X ,^ [or .473] = 33.11 Gm., the amoimt of Iodine for 1 pint. 3. Calculate the quantities for I gallon of soap lini- ment from the following formula : — soap, 60 Gm. ; camphor, 45 Gm. ; oil rosemary, 10 c.c. ; alcohol, 725 cc. ; water, q. s. ad 1000 c.c. 4. Calculate the formula for i Kg. of tooth powder from the following formula : — Powdered soap, 4 oz. ; precip. chalk, j/2 lb. ; camphor, 30 gr. ; vanillin, 5 gr. ; oil rose, 8 TTL ; pow- dered sugar, 2 oz. ; magnesium carbonate, q. s. ad i lb. Solution. — I Kg. = 2.2 lb. Hence each quantity is to be multi- plied by .2^%- that is, by 2.2. 4 oz. X 2.2 = 8.8 oz. = 249,48 Gm. % lb. X 2.2 = 1.1 lb. = 500 Gm, 30 gr. X 2.2 = 66 gr. = 4.29 Gm. 5 gr. X 2.2 = 11 gr. = .715 Gm. 8 ni X 2.2 = 17.6 Vl = 1.08 cc. 2 oz. X 2.2 = 4.4 oz. = 124.74 Gm. Mag. carb. q. & ad 2.2 lb. = q. s. ad 1 Kg. Note. — Remember that the product is always in the same unit as the multiplicand. 5. The National Formulary gives the following formula for Dewee's Carminative: — ^magnesium carbonate, 50 Gm. ; tr. asaf oetida, 75 cc. ; tr. opium, 10 cc ; sugar, 100 Gm. ; water, q. s. ad lOOO c.c (a.) Calculate quantities for 120 cc; (Ey for 4 f||; (c.) for }4 Cong. ; (d.) for 2 L. ; (e.) for 3750 cc Reducing and Enlarging Formulas 65 6. The formula for compound syrup of squill is: — fl. ext. squill, 80 c.c. ; fl. "ext. senega, 80 c.c. ; antimony and potassium tartrate, 2 Gm. ; purified talcum 20 Gm. ; sugar, 750 Gm. ; water, q. s. ad 1000 c.c. (a.) Calculate quantities for i O. ; (b.) 180 c.c. ; (c.) for I Cong.; (d.) for 5850 c.c; (e.) for 2 O. 6 f §. 7. A formula for deep blue show globe color is : — cop- per sulphate, 15.5 Gm. ; water, 250 c.c. Make solution, to which add — ammonia water, 75 c.c. ; and then enough water to make 2 L. (a.) Calculate formula for 5780 c.c, (b.) for 2 gallons. Ans. Copper sulphate, 58.67 Gm., water, 946 c.c, ammonia water, 283.9 c.c ; and enough water to make 2 gallons. Calculating Definite Weight from Parts by Weight. *Tarts-by-weight-formulas" are given in the U. S. Pharmacopoeia of 1880, in the latest edition of the German Pharmacopoeia (1900), in the National Formulary, and in many scientific publications. The advantage of such a formula lies in the fact that a part may mean a pound, an ounce (av.), an ounce (apoth.), a grain, a Gramme, a Kilogramme, — ^in short, any convenient unit of weight. But for formulas for liquids, "parts by weight" were never popular in the United States, because of the established custom in our country of measuring liquids. Problems. 8. Calculate a formula for 5 lb. of salicylated talcum from the following formula of the German Pharmacopoeia of 1900 : — salicylic acid, 3 parts ; wheat starch, 10 parts ; talcum, 87 parts. Solution. — A part in this case is assumed to be a pound. Then the general rule, to multiply each quantity by the fraction of which the amount wanted is the numerator, and the amount the formula makes is the denominator, is applied. The amount wanted is 5 lb.; the formula makes 100. Hence each quantity is multiplied by jhs or -05. Thus we have — Salicylic acid 3 lb. X .05 = .15 lb. Wheat starch 10 lb. X .05 = .5 lb. Talcum 87 lb. X .05 = 4.35 lb. 66 Pharmaceutical Arithmetic The fractions of pounds should, of course, be reduced to lowef units. 9. Suppose 500 Gm. of the salicylated talcum are wanted. Each part is assumed to be a Gramme. Then — 500 3 Gm. X-^= 15 Gm. 10 Gm. X 6 =50 Gm. 87 Gm. X 5 = 435 Gm. 10. Suppose 480. gr. are wanted for a prescription. Each part is assumed to be a grain. Then — 3 gr. X m = 14.4 gr. 10 gr. X 4.8 = 48. gr. 87 gr. X 4.8 = 417.6 gr. 11. The German Pharmacopoeia of 1900 gives the fol- lowing formula for powder of magnesia and rhubarb: — magnesium carbonate, 50 parts ; eleosaccharate of fennel oil, 35 parts; rhubarb, 15 parts, (a.) Calculate quantities for 600 Gm. of the powder; (b.) for 2 Kg. ; (c) for 2 lb. ; (d.) for 8 5 ; (e.) for 8 oz. ; (f.) for 300 gr. ; (g.) for 6 5 2 3 i 9 10 gr. Ans. (a.) (d.) (e.) Mag. carb. 300 Gm. 4 5 4 oz. Eleosacch. of fennel 210 Gm. 2 5 6 3 1 3 4 gr. 2 oz. 350 gr. Rhubarb 90 Gm- 1 5 1 5 1 9 16 gr. 1 oz. 871/2 gr. 12. Recipe for polishing paste: — oxalic acid, i part; Jewelers' rouge, 16 parts ; rotten stone, 20 parts ; palm oil, 59 parts ; petrolatum, 4 parts, (a.) Calculate formula for i lb.; (b.) for I Kg. ; (c.) for 12 oz. ; (d.) for 400 Gm. CHAPTER V. Proportions. To solve the problem — If 20 lb. of tartaric acid cost $7, liow much will 32 lb. cost ? — we reason as follows : — i lb. of acid will cost $7.00 h- 20 = $.35 ; and 32 lb. will cost 32 times as much as i lb., hence $.35 X 32 = $11.20. We may arrive at the same answer by an operation based upon a different line of reasoning, thus: — Price and quantity in this problem bear such a relationship to each other that they increase and decrease in the ratio. If, for instance, the quantity is doubled, is raised to 40 lb., the price must be doubled likewise, must be raised to $14.00. Notice that the increase is not dollar per pound, but at the same rate in case of the dollars as in case of the pounds. If the quan tity is reduced to one-fourth, to 5 lb., the price becomes $7.00 -T- 4 = $1.75. Then if we can find the ratio of 20 lb. to 32 lb., we have also the ratio of $7.00 to the answer sought. When this definite relationship exists between values — when they increase and decrease in the same ratio-=— the val- ues are said to be proportional to each other. Thus quanti- ties and prices are proportional. The statement that 20 lb. bears the same relation to 32 lb. as $7.00 bears to $11.20, is called a statement of a propor- tion. The statement may be more concise as follows. — 20 lb. is to 32 lb., as $7.00 is to $11.20. For is to, it is customary to use the sign of ratio, the colon [ :] ; and for as, the sign of proportion, the double colon [ ::]. Thus — 20 lb. : 32 lb :: $7.00 : $11.20. is to as is to Each one of the members of the proportion is called a term. Terms are numbered from left to right. Thus in the preceding example 20 lb. is the first term, 32 lb., the second term, $7.00, the third term, and $11.20, the fourth. The first two terms constitute the first couplet, and the last two, the second couplet. Pharmaceutical Arithmetic The first and last terms are called the extremes; the second and third, being in the middle, are called the means. 1st couplet 2nd couplet 20 lb. 1st term 32 lb. 2nd term $7.00 3d term $11.20 4th term means. extremes It will be seen that the product obtained by multiplying the two means, and the product obtained by multiplying the two 'octremes are the same. 20 X 11.20 = 224.00 32 X 7.00 = 224.00 It follows that the product of the extremes divided by either mean will give the other mean. And the producl of the means divided by either extreme will give the other extreme. 224 -f- 20 = 11.20; 224 -f- 11.20 = 20 224 -f- 32 = 7.00; 224 -V- 7.00 = 32 Hence if three terms are known, the fourth one may be calculated. The missing term may be any one of the four, as fol- lows: 1. If 20 lb. of acid cost $7.00, how many lb. of add may be bought for $11.20? 2. If 32 lb. cost $11.20, how many lb. could be bought for $7.00? 3. If 20 lb. cost $7.00, how much would 32 lb. cost ? 4. If 32 lb. cost $11.20, how much would 20 lb. cost? The preceding four problems may be stated thus : I. x : 20 lb. : $11.20 : $7.00 2. 32 lb. X :: $11.20 : $7.00 3. 32 lb. : 20 lb. : : X : $7.00 4. 32 lb. . 20 lb. . : $11.20 X Proportions 69 And solved thus: 1 and 4. Multiply the two means, and divide by the given extreme. 2 and 3. Multiply the two extremes, and divide by the given mean. It is, however, simpler and less apt to lead to error, if the proportion he always so stated that the missing term is the fourth. This can be done by applying the following rules: 1. Let the 3d term be that number which expresses the same kind of value as will be expressed by the answer sought. 2. Then decide whether the answer sought is to be greated or smaller than this 3d term. If greater, place the greater of the two remaining terms as the 2nd term; if smaller, place the smaller of the two remaining terms as the second. 3. The remaining known term place as the first. Note. — ^The Ist and 2nd terms must be in the same unit of value, in order that the ratio of the abstract numbers may also be the ratio of the denominate numbers. Since the missing term is one of the extremes, it is obtained by multiplying the two means — the 2nd term and the 3d — and dividing the product by the ist term*. Note. — The answer will be in the same unit as the 3d term. If it is to be in a different unit it must subsequently be reduced to that unit. Problem. — If 437.5 gr. (i oz.) of phenacetine cost 95c, how much will 120 gr. (2 3) cost? Solution. — ^Th« three known terms are — 95c, 437.5 gr., 120 gr. The answer will express cost. i. e., money-value. Therefore the known term expressing cost should be the third term: : : : 95c : x (Answer.) Since the quantity for which the cost is to be found is less than the quantity for which the price is given, the answer will be smaller than the third term. Hence the smaller of the two remaining terms is to be the second: : 120 gr. :: 95c : x As but one known term remains unplaced, this must be the first; •This is identical with dividing the known term «f the second couplet by the ratio of the 1st term to the 9n/l, 70 Pharmaceutical Arithmetic 4S7S gr. : 120 gr. : : 95c : x The 1st and 2d terms may now be treated as abstract numbers: 95c X 120 = 11400c 11400c -^ 437.5 = 26c. (Answer.)* Solving the problem by analysis, we would reason thus : —if 437.5 gr. cost 95c, I gr. will cost 95c -v- 437.5 ; and 120 gr. will cost 120 times as much as i gr. In this case the division is carried out first, and the quotient is multiplied, while in the proportion the multiplication is carried out first and the division last. One of the fundamental rules of arithmetic is that multiplication and division may be carried out in any order. So a proportion could be com- pleted by dividing the first term into the second or third, and then multiplying the remaining known term by the quo- tient. But it is preferable to carry division out last — not only in proportions, but in all problems — for the reason that the quotient may be burdened with an interminate decimal, the rounding off of which occasions an error, an error that a subsequent multiplication would magnify. Problems. I. If cocaine costs $6.50 an oz., how much will 3 3 cost? Solution. — The three known terms are: $6.50, 1 oz., ond 3 3. The answer is to express cost. Hence — :: $6.50 : x The Ist and 2d terms must be in the same unit. In this prob- lem the 1 oz. and the 3 3 are best reduced to gr. 1 oz. = 437.5 gr., and 3 3 = 180 gr. [If the cost were known per 5, it would be simpler to reduce to 3.] •Cancellation. The figuring may frequently be lessened by cancellation. That is, by divid- ing the first term by a certain number, and then one of the other t&ma by the same number, which proceas in no way changes the ratios. Example,— 3 91 m : m : : 35^ : x The first term and also the third may be divided by 10 by cancelling the ciphers. Then the first term and the second may each be divided by 9. The work has now been sunplifled to 91 X 35 -i- 3. In proportions stated so that the unknown term is the fourth, the second and third terms are multiplied, one with the other, and hence cannot be cancelled against each other. Proportions 71 The quantity for which the cost is to be found is smaller than *he quantity for which cost is given. Hence the answer will be smaller — 437.5 gr. : 180 gr, : : $6.50 : x $6.50 X 180 = $1,170.00; $1,170.00 -j- 437.5 = $2.67. 2. (a.) If cocaine costs $6.50 an oz., how much will 2 5 I 9 5 gr. cost? (b.) How much will 2 Gm. cost? (c.) How much will 3.256 Gm. cost? Ans. (c.) 74c. 3. (a.) If chloral hydrate costs $1.30 a lb., how much will I 5 cost? (b.) How much will 10.560 Gm. cost? 4. (a.) If cream of tartar costs 60c per Kg. (KjIo.), how much will i lb. cost?- (b.) How much will i § cost? (c.) How much will 2^ oz. cost? (d.) How much will 2 5 2 3 2 9 2 gr. cost? Ans. (a.) 27c. 5. If homatropine costs $6.00 a Gm., how many gr. can be bought for $1.75? Ans. 4y2 gr. Remarks. — ^The answer is to express quantity ; hence i Gm. is the 3d term. The answer will be in Gm., and must be reduced to gr. 6. If homatropine costs $6.00 a Gm., how much would 2 gr. cost? Ans. Nearly 78c. 7. If 15 gr. of heroin cost i8c, how much is I oz. worth ? 8. (a.) If alcohol costs $2.40 a gallon, how much does a Liter cost? (b.) How much would 225 c.c. cost? (c.) How much would 2 O. 6 f 5 cost? (d.) How much can be bought for $100? 9. Glycerin cost 23c per lb. Its sp. gr. is 1.25. How much does it cost a Liter ? Solution. — 1 L. = 1000 mils ; 1000 X 1.25 (sp. gr.) = 1250 Gm. 1 lb. = 453.6 Gm, Then 453.6 Gm : 1250 Gm. : : 23c : (x) 67.8c Note. — Quantity is proportional to price; and the quantity may be expressed in weight or volume, and in any tmit of these, pro- vided the 1st and ad terms are in the same unit. So this and sim- ilar problems may be solved in sevwal different ways. 10. Oil of lemon costs $1.25 a lb., and has a sp. gr. of J855. How much does it cost per f^? Ans. 7c. 11. (a.) Mercury costs $1.50 per Kg. Taking its sp. 72 Pharmaceutical Arithmetic gr. as 13.5, how much would 4 f § cost? (b.) How much would 12 c.c. cost? (c.) How much would 2 f 3 45 TT], cost? 12. Alcohol costs $2.40 a gal., and has a sp. gr. of .81. How much does it cost per lb.? Ans. 35.6c. 13. Castor oil costs 14c. a lb., and has a sp. gr. of .95. How much does it cost a gal? Ans, $1.10. 14. Chloroform costs 60c. a lb., and has a sp. gr. of 1.48. How many f '^ can be bought for 75c? Ans. 13 f '• 15. If 4 Gm. of mustard will make 60 square centi- meters of mustard paper (plaster), how much mustard should be spread on a plaster 3 in. by 4 in.? Solution.— 3 in. by 4 in. = 7.6 cm. by 10.2 cm.; and 7.6 cm. X 10.2 cm = 77.52 sq. cm. Then 60 sq. cm. : 77.52 sq. cm. : : 4 Gm. : x (5.166 Gm.) 16. If I gal. of paint costs $1.35, and will cover 500 sq. ft., what would be the cost of 3 coats for a house 18 ft. high 30 ft. broad, and 45 ft. long? Remarks. — ^The relationship between price and quantity, or be- tween quantity of paint and amount of surface to cover, as in problem 16, is clearly seen, and that these values are proportional IS quite obvious. But in many pharmaceutical problems the relation- ship is not so apparent, or, may appear to exist when it does not. In the appendix is given a list of proportional values, which list should be frequently consulted. Certain values bear such a- relationship to each other that one increases in the some ratio in which the other decreases. Values proportional in this sense are said to be inversely proportional. Bee Chapter VIL TRADE DISCOUNT. Problems involving interest, profit and loss, premiimis, discount, as well as all other percentage problems, may be solved without the use of specific rules, if the general directions for stating pro- portions are observed. It should be remembered, however, that if several discounts are given, the second is based not upon the regular selling price, but upon the price obtained by deducting the first discount; that the third is based upon the price obtained by deducting the first and second discounts; etc. Problem. — Certain apparatus is quoted at $120.00 per gross, with 40%, and 10% off, and an additional discoimt of 3% for cash. What is the net cost? Solution.— 40% of $120.00 = $48.00; $120.00 mmus $48.00 8=^2.00. 10% of $72.00 = $7.20; $72.00 minus $7.20 = $64.80 3% of $64.80 = $1.94; $64.80 minus $1.94 = $62.88 (Ans.) CHAPTER VL PERCENTAGE PROBLEMS. Percentage — abbreviated percent., and indicated by the symbol % — means parts per lOO parts. Opium, said to contain 14% of morphine, contains 14 parts of the latter to 86 parts of other constituents, making 100 parts in all. The amount of a constituent in a drug, the amount of component in a mixture or in a preparation, the amount of dissolved substance in a solution, — all these amounts are conveniently expressed in percentages ; that is, they are con- veniently adjusted to the scale of 100*. This being- true, problems involving percentages are of frequent occurrence in every-day work. In such problems the following factors come into play : 1. The amount of drug, mixture, preparation, or so- lution, or whatever is to represent the 100 parts. 2. The amount of constituent or ingredient., 3. The amount of constituent or ingredient as ex- pressed in percentage. 4. The amount of drug, mixture, etc., expressed in percentage. This is always 100, and is never sought as an answer. 5. The amount of diluent or solvent. For sake of brevity let us designate drug, mixture, preparation, or solution by M, standing for mixture; con- stituent or ingredient by C, standing for constituent; per- centage by its customary symbol, % ; and diluent or solvent byD. In percentage problems there are four cases possible: (i) M may be sought, the other factors being known; (2) •In order that comparisons may be readily made — ^made by inspection— mich data should be reduced to some common standard or scale; and the number lOb is selected, rather than 12, Hi, or any other numlvH-, for arithmetical reasons. 74 Pharmaceutical Arithmetic C may be sought; (3) % may be the element to be calai- lated ; (4) the amount of diluent or solvent may be sought. It should be remembered that percentages in pharma- ceutical or chemical problems refer to parts by weight, un- less the contrary is expressly stated. Now, if percentages stand for parts by weight, they are proportional to weights of M and of C, and any one of the first three factors — M, C, or % — may be calculated by proportion. D is the difference between C and M, and is found by subtracting C from M. See page 79. Examples. 1. Suppose that 6 Gm. of opium on analysis are found to contain .585 Gm. of morphine, and the morphine-strength of the opium is to be expressed in percentage. Solution: — The three known terms are: — 6 Gm. (amount of drug = M), .585 Gm. (amount of morphine, i. e., constituent = C), 100 per cent, (per cent of M)*. The missing term is per cent, of C. ^ Bince the answer is to be in per cent., the known per cent., 100 per cent. (M Is always assumed to be 100 per cent.), must be the third term. : :: 100 per cent. : x The answer is to be smaller than 100 per cent, because the amount of C. (.585 Gm.) Is smaller than the amount of M (6 Gm.). Or, to generalize, the per cent of C must always be less than 100, because a part Is less than the whole. Therefore the smaller of the two remaining known terms is the second, and the larger, the first C Gm. ; .585 Gm. : : 100 per cent : x X = 9.75 per cent. 2. How much morphine is present in 6o Gm. of opium having a morphine-strength of 12% ? The three known terms are: — 60 Gm. (amount of M), 12 per cent, (per cent, of C), and 100 per cent, (per cent, of M). •Remember that M alwajv Btands for the medicinal substance or preparation in its totality. So in this and the immediately succeeding problems it stands for opium, and not tor morphine, although the latter word begins with m. The choice of drug and constituent might appear unfortuiiate; but the selection of a constituent with a name beginning with m was premeditated, the object being to point out that the sjmboL) M and C are general, and hare no relation* «Up to the names of ^eciflc drugs, or of preparations, or of constituents. I Percentage 75 The answer is to express weight. Therefor©—* J :: 60 Gm. : x The answer is to be smaller than 60 Gm., because 12 per cent, is less than 100 per cent. And because C, the part must weigh less than M, the whole. Then the smaller remaining term (12 per cent.) must be the second. Thus — 100 per cent : 12 per cent ; : 60 Gm. : x X = 7.2 Gm. 3. If a certain opium has a morphine-strength of 14%, how much of the opium would contain just 5 Gm. of morphine ? The known terms are: — 100 per cent, (per cent, of M), 14 pel cent, (per cent, of C), and 5 Gm. (wt. of C). The term sought is wt. of M. Since the answer is to be in weight, the known weight is thfl third term. t ;: 5 Gm. : x The answer is to be greater, because 100 per cent, is greatex than 14 per cent, and because the whole must be greater than the part. Hence — 14 per cent. .: 100 per cent. : : 5 Gm. : x X = 35.714 Gm. While th6 rules on page 69 for stating proportions are always applicable, and cannot mislead, it is well to know that in percentage proportions correctly stated, M and C al- ternate ; that is, the term giving % of M and the one giving weight of M are never side by side, but have another term between them,— which is true likewise, for the two terms giving respectively weight and % of C. Observe the three proportions just g^ven. I. 6 Gm. : .585 Gm. : : 100% : (x) 9.75% wtofM. wt.ofC. %ofM. %ofa 76 Pharmaceutical Arithmetic 2. 100% : 12% : : 60 Gm. : (x) 7.2 Gm. % of M. % of C wt. of M. wt of C 3. 14% : 100% : : 5 Gm. : (x) 35.714 Gm. % of C % of M wt. of C wt. of M.. N. B. — ^A similar alternation occurs in all proportions in which the two values are directly proportional ; if they are inversely proportional [see chapter VIII] the alternation does not occur. Other Arithmetical Processes. Problem 1 may be solved also by reasoning as follows: If 6 Gm, of opium contain .585 Gm. of morphine, 1 Gm. of opium will con- tain .585 Gm. -7- 6 = .0975 Gm.; and 100 Gm. will contain 100 time* as much as 1 Gm., hence [.0975 Gm X 100 = ] 9.75 Gm. Consequently the answer is, 9.75 per cent. Formulating this we have — C X 100 % = (C -^ M) X 100 or, % = "^ M Problem 2 is solved as follows: If the morphine strength of the opium is 12 per cent., each 100 Gm. of the latter contain 12 Gm. of morphine, and 1 Gm. of opium contains 12 Gm. -r- 100 = .12 Gm. Then 60 Gm. contain 60 times as much as 1 Gm., that is [.12 Gm. X 60= ], 7.2 Gm. Expressing the above process in a formula, we have— % X M C = (% -^ 100) X M or, C = 100 Problem 3: If the morphine-strength of the opium is 14 per cent., 100 Gm. of the latter contain 14 Gm. of morphine. Then 1 Gm. of morphine will be contained in [100 Gm. -t^ 14 Gm. =] 7.1428 Gm. of opium; and 5 Gm. of morphine will be contained in 7.1428 Gm. X 5 = 35.714 Gm. This expressed in a formula gives: 100 X C M = (100 -T- %) X C or, M = % The same result may be obtained as follows: In ICO Gm. of opium there are 14 Gm. of morphine. In 1 Gm. of opium there are Percentage jy 14 Gm. -r- 100 = .14 Gm. of morphine. Then the answer will be: as manv Gm. of opium as .14 Gm. is contained in 5 'Gm. = 35.714 Gm. Formula' — M = C -^ % -T- 100. Since dividing the divisor Is the same as multiplying the divi- dend, 100 X C = C ^ % ^ 100, % and the two processes must give the same answer; — must both be correct. Advice to Students. The formulas in the preceding paragraphs should not be memorized, and followed blindly. Students who cannot solve percentage problems by a process of reasoning, are strongly advised to avoid all formulas and rules (which are usually remembered imperfectly), and to solve all such prob- lems by proportion. The principles underlying proportions are soon mastered by the average students; and by the ex- ceptional ones, who are absolutely unmathematical, the pro- cess may be followed empirically without danger of error, as, in proportions, no matter what the case — no matter which factor is missing — the rules to be followed are the same. (See page 69). Problems. 4. A certain drug contains 17% of soluble matter, hence will yield 17% of extract. How much extract could be obtained from 500 Gm. of the drug ? Ans. 85 Gm. 5. How many lbs. of the same drug would be required to make 5 lbs. of extract? Ans. 29^3^ lb.., or 29.41 lb. 6. What is the percent, of extractable matter in a certain licorice root, 500 Gm. of which yield 90 Gm. of extract ? 7. Diluted hydrocyanic acid, U. S. P., contains 2% of absolute hydrocyanic acid (HCN), the remainder being water. How much absolute acid in 450 Gm. ? Ans. 9 Gm. 8. Official nitric acid contains 68% of absolute nitric 78 Pharmaceutical Arithmetic acid (HNO3), the remaining 32% being water. How many gr. of absolute HNO, in 400 gr. ? Ans. 272 gr. 9. How many Gm. of iodine in 300 Gm. of a 5% solu- tion? 10. How much iodine should be weighed out to make I Kg. of a 5 % solution ? II.. How many Gm. of 5% soda solution can be made from 400 Gm. of soda? 12. What is the percentage strength of a solution of homatropine made by dissolving 15 gr. in enough water to make 400 gr. ? Ans. 3.75%. 13. What is the percentage strength of a solution of mercuric chloride made by dissolving 60 gr. of the latter in enough water to make 5,000 gr. ? 14. What is the percentage strength of a solution of sodium chloride made by dissolving 65 gr. of the salt in 400 gr. of water? Solution. — 65 gr. in 400 gr, would make 465 gr. of solution. The three known terms are: [400 + 65 =] 465 gr., 6C gr., 100 per cent. Percentage is asked for. Hence — : : : 100% : x. The answer is to be less than 100%. Hence — 465 gr. : 65 gr. :: 100% : (x) 13.9% 15. What is the percentage strength of a solution of carbolic acid in glycerin made by mixing 350 Gm. of the latter with 200 Gm. of the acid? 16. Boric acid is soluble in 18 parts of water. What is the percentage strength of a saturated solution? Solution.— 1 part of acid + 18 parts of water = 19 parts of solution. Then 19 parts : 1 part : : 100% : x (5.26%). 17. Potassium chlorate is soluble in 16 parts of water. What is the per cent, strength of a saturated so- lution ? 18. One part of water will dissolve 1.4 parts of potassium iodide. What is the per cent, strength of a sat- urated solution? Ans. 58.3%. 19. One part of water will dissolve -^ part of po Percentage 79 tassium permanganate. What is the per cent, strength of a saturated solution? Ans. 6.25%. 20. What percentage of sugar in a syrup made from I lb. of sugar and ^ lb. of water? Ans. 66 2-3%. Cai^ui^ting Amount dp Dii,uent or Soi,vent. To make 100 Gm. of a 15% solution of salt we may proceed in two ways : we may weigh out 15 Gm. of salt, and then add enough water to bring the total weight to lOO Gm. ; or, we may add to the 15 Gm. of salt 85 Gm. of water, pre- viously weighed. In the first method there is a possibility of adding too much water; and as solution begins at once, the excess of water cannot be removed without removing some salt also. For this reason the second method is pre- ferred. 21. How many lbs. of sugar, and how many of water, are required to make 12 lbs. of syrup having a sugar-content of 65%? Solution.— 100 per cent. : 65 per cent. : : 12 lbs. : x X = 7.8 lbs. (wt. of sugar). Then 12 lbs. — 7.8 lbs. = 4.2 lbs. (wt. of water). 22. In how many Gm. of water must 15 Gm. of co- caine hydrochloride be dissolved to make a 4% solution? Solution. — 4 per cent. : 100 per cent. :: 15 Gm. : x X = 375 Gm. (wt. of solution to be made). Then 375 Gm. (weight of solution) minus 15 Gm. (wt of cocaine) = 360 Gm. (wt. of water). 23. How many Gm. of magnesium sulphate, and how many Gm. of water, are required to make 600 Gm. of 28% solution? Ans. 168 Gm. mag. sulph. and 432 Gm. water. 24. How many oz. of carbolic acid, and how many of petrolatum, must be used to make 32 oz. of carbolated pe- trolatum containing 3% of carbolic acid? 8o Pharmaceutical Arithmetic 25. How many Gm. of water must be used to dissolve 4 Gm. of mercuric chloride to make a. -^ >% solution ? 26. A certain nasal spray consists of liquid petrolatum, with 6% of menthol, 5% of eucalyptol, and 2% of thymol. How much must be used of each of the ingredients to make 150 Gm? Solution.— 100% : 6% :: 150 Gm. : x X = 9 Gm. (am't of menthol). 100% : 5% :: 150 Gm. : x X = 7.5 Gm. (am't of eucalyptol). 100% : 2% :: 150 Gm. : x X = 3 Gm. (am't of thymol). Then 9 Gm. + 7.5 Gm. + 3 Gm. = 19.5 Gm. (am't of active ingredients) ; And 150 Gm. (am't of spray) — 19.5 Gm. (am't of active in- gredients) = 130.5 Gm. (am't of liquid petrolatum). 2y. ^ Iodoform gr. xxx Boric acid 3 j Naphthalin 3 j Make a fine powder, and flavor with 2% of oil of ber- g^mot. How much of the latter must be used ? Ans. 3 gr. 28. How much oil, acacia and water are required to make the nucleus of 120 c.c. of a 40% emulsion. Sp. gr. may be ignored. Solution. — ^By the continental method the nucleus is made of 4 parts of oil, 1 part of acacia, and 2 parts of water. 40 per cent of 120 c.c. = 48 c.c. (amount of oil). Then If 4 parts = 48 c. c. (amount of oil), 2 parts = 24 c, c. (amount of water), 1 part = 12 Gm. (amount of acacia). This nucleus is then to be diluted to 120 c.c, making a 40 per cent, emulsion. Percentage Problems Involving Conversion ^om Onb System of Weight to Another. 29. How many Gm. of 4% solution can be made with I oz. of cocaine hydrochloride? Ans. 708.75 Gm. Percentage 8i 30. What is the per cent, strength of a morphine sul- phate solution made by dissolving ys oz. in enough water to make 100 Gm? Ans. 3.543%. 31. To make 5 lbs. of 5% solution of soda, how many Gm. of the latter must be used? 32. How many Gm. of 2% atropine sulphate solution can be made with i 3 I 9 10 gr. ? Problems Involving Also Reduction op Volumes to Weights, or Weights to Volumes. Since in these problems — and always when the con- trary is not expressly stated — ^percentages stand for parts by weight, they are proportional to weights, but not to volumes. If quantities [of M or of C] are given in volumes, these must be reduced to weights before employing the quantities as terms in proportions. If the answers (4th terms) of these proportions represent quantities, these will be in weight — not in volume. So a conversion to volume must follow whenever the answer is to express volume, 33. Official nitric acid has a sp. gr. of 1.403, and con- tains 68% of absolute acid. How many Gm. of abs. acid in 2 L. of the official acid ? Solution,— 2 L, = 2000 c.c. 2000 X 1,403 (sp. gr.) = 2806. Hence 2000 c.c. = 2806 Gm. Then 100% : 68% : : 2806 Gm, : x X = 1908 Gm, (wt, of abs. acid). 34. Official solution of potassa has a sp, gr, of 1,046, and contains 5% of potassa. How many f^ of solution could be made from 20 Gm. of potassa ? Solution.— 5% : 100% : : 20 Gm, : x X = 400 Gm. (weight of solution), 400 -T- 1,046 (sp, gr.) = 382.4 (vol, in c.c.) 382.4 -T- ''9.57 (c.c. in 1 fS) = 12.93 fj 35. Syrup contains 85 Gm. of sugar in 100 mils, and has a sp, gr. of 1,313. What per cent, of sugar does it contain ? 83 Pharmaceutical Arithmetic Solution.— The sp. gr. being 1.313,100 mils of syrup weigh 131.3 Gm. Then— 131.3 Gm. : 85 Gm. : : 100% : Z X = 64.7% 36. Stronger ammonia water has a sp. gr. of .897, and contains 28% of ammonia. How many Gm. of the latter in I pint of the stronger water? Ans. 118.8 Gm. S7. Official sulphuric acid has a sp. gr. of 1.826, and contains 92.5% of absolute acid, (a.) How many gr. of absolute acid in i f 5 of official add? (b.) How many Gm. of abs. acid in 2 L. of official acid? Ans. (a.) 769.7 gr. 38. A solution of ferric citrate has a sp. gr. of 1.25 ; and i L. yields 531.25 Gm. of ferric citrate (scale salt). What per cent, of ferric citrate in the solution ? 39. In a certain chemical reaction 5 Gm. of absolute hydrochloric acid (HCl) is required. To supply it how many c.c. of official acid (sp. gr., 1. 158, absolute-acid-con- tent, 31.9%) must be used? Ans. 13.47 c.c. 40. 30 c.c. of ether, sp. gr., .716, is dissolved in enough alcohol to make 100 c.c. of solution, the latter having a sp. gr. of .8. What per cent, of ether (by v/eight) in the pre- paration ? 41. Lead subacetate solution has a sp. gr. of 1.235, and contains 25% of lead subacetate. On diluting 3 c.c. of this solution to 100 c.c, diluted solution of lead subacetate (U. S. P. 1890) is obtained. How much lead subacetate does the latter solution contain in each f §? Solution. — I f 5 = 29.57 c.c. If 100 c.c. of the diluted solution contain 3 c.c. of the strong solution, i c.c of the diluted solution will contain 3 c.c. -^ 100 = .03 c.c. of the strong; and 29.57 c.c. of the diluted solution will contain .03 c.c. X 29.57. = .8871 c.c. of the strong solution. The sp. gr. of the latter being 1.235, the .8871 c.c. weighs [.8871 X 1.235 = ] 1-0955 Gm. Of this 25% is lead subacetate. Then— 100% : 25% :: 1.0955 Gm. • (x) -274 Gm. 42. Solution of ferric sulphate has a sp. gr. of 1.432, ,and contains 36% of ferric sulphate. How many g^. of Volume Percentage 83 ferric sulphate in i f 5 of a diluted solution, containing 20 C.C. of the official solution in 100 c.c. ? Ans. 48 gr. Volume Percentage. A volume percentage solution is one having in 100 volumes a definite number of volumes of constituent, the latter necessarily being a liquid. Thus, alcohol, U. S., con- tains 94.9% by volume of absolute ethyl hydroxide, the re- maining 5.1% by volume being water. Diluted alcohol, U. S., contains 48.9% by volume of ethyl hydroxide, and 51.1% by volume of water. Notice that it is expressly stated that the percentage is by volume. When it is not so stated, weight percentage is always understood. , Since percentages in volume percentage stand for vol- umes, they are proportional to volumes just as weight per- centages are proportional to weights. In other words, vol- ume percentages are proportional to L., c.c, f §, TTt, etc., but not to weight units. Problems. 44. In I gallon of an alcohol, containing 94% by vol, of ethyl hydroxide, there are how many f § of the latter ? Ans. 120.3 f S- 45. In 5 L. of diluted alcohol, containing 48.9% by vol. of ethyl hydroxide, there are how many c.c. of the latter? 46. A certain alcohol contains 6 O. 8 f ^ of abs. ethyl hydroxide in each gallon. What % by vol. of ethyl hydrox- ide does the alcohol contain? Ans. 81.25% by vol. 47. A certain menstruum is to contain 15% by volume of glycerin. How much of the latter should a pint con- tain ? Ans. 2 f 5 3 f 3 12 TT].. Calculating Volume Percentage to Weight Percent- age AND Vice Versa. If the sp. gr. of the constituent is known, and also the sp. gr. of the solution, volume percentage may be calculated to weight percentage, — and vice versa. §4 Pharmaceutical Arithmetic 48. An alcohol, sp. gr., .82, containing 94% by volume of ethyl hydroxide contains what % by weight ?* Solution. — ^The ap. gr. of the alcohol [solution] is given as .82; that of ethyl hydroxide [constituent] as .79. If the sp. gr. of the alcohol is .82, 100 c.c. of it weigh 82 Gm. If the sp. gr. of the ethyl hydroxide is .79, 94 c.c. of it would weigh [94 X .79 =] 74.26 Gra. Then— 82 Gm. : 74.26 Gm. : : 100% : (x) 90.56% wt of M : wt. of C : : % by wt of M : % by wt of C 49. An alcohol is 91% strong by weight, and has a sp. gr. of .82. Calculate strength in volume percentage.* Solution.— 100 Gm. of alcohol = 100 t- .82 = 121.9 c.c. 91 Gm. of ethyl hydroxide = 91-7- .79 = 115 cc Then 121.9 cc. : 115 cc. : : 100% : (x) 94 -f- % vol. of M : vol. of C. : ; vol. % of M. : vol. % of 0. 50. The sp. gT. of ethyl hydroxide being .79, what is the wt. per cent, strength of an alcohol of 66% by vol., and having a sp. gr. of .9? 51. An alcohol has a sp. gr. of .8719, and contains 70% by wt. of ethyl hydroxide [sp. gr. .7935]. What % by vol. of the latter does it contain? [See alcohol table, U. S. P.] Note. — ^By the Internal Revenue Office strength of alcoholic liquids is expressed by degrees proof. An alcoholic liquid con- taining 50 per cent, by vol. of absolute ethyl hydroxide is said to be 100 proof. An alcohol, 94 per cent by vol. strong, is said therefore to be 188 proof. In short — degrees proof -4- 2 = strength in per cent, by vol.; and per cent by vol. x 2 = de^ grees proof. 52. What is the alcohol strength of brandy marked no proof? Solution. — no -f- 2 = 55% by vol. 'Answers to 48 and 49 will harmonize if specific srravities are not rounded ofl^ M in the above solatioiu Percentage — Sp. Gr. Unknown 85 To Calculate Amount op Constituent When the Sf* Gr. op Solution is Unknown. Percentage, unless the contrary is expressly stated, re- fers to parts by weight. If the amount of constituent is to be calculated from the amount of solution, the weight o£ the latter must be known,— or must be calculated from the volume and the sp. gr., as in the problems on page 81. In practice, however, the pharmacist is often called upon to dispense a definite volume of a solution of specified percentage strength but of unknown sp. gr. For instance;^ I f o of a 4% solution of cocaine hydrochloride may be wanted. Now, the error due to the assumption that the sp. gr. of such a low-strength solution is i.ooo, would be very slight, and would be considered negligible by most physi- cians. But the doctrine of permissible inaccuracy is a dangerous one; ajid the student is not advised to take "short cuts" which affect the strength of a medicinal prepara- tion. A solution of exact strength can be made as follows : Use as large a volume of solvent as volume of product desired; and base calcidation of atnount of constituent upon the weight of this volume of solvent. The constituent will augment the volume somewhat; if an exact volume is wanted, this may be measured out, and the excess either saved or rejected — according to i*ts stability, commercial value, etc. To illustrate, let us revert to the problem just g^ven, calling for i f 5 of a 4% solution. Calculation. — Since the solution is 4 per cent, strong, the solvent represents 96 per cent.; and if -we take 4 parts by weight of con- stituent as often as we have 96 parts by weight of solvent, our finished product will contain 4 parts in 100 parts, i, e., will be ex- actly 4 per cent, strong. Accordingly. 96% : 4% : : 454.6 gr.. (wt. 1 fS "ssrater) : x X = 18.94 gr. 86 Pharmaceutical Arithmetic Thus by measuring out i f § of water, and dissolving In it 18.94 gr. of constituent, we get a trifle (about 9 TTt) over I f o of solution, exactly 4% strong. The superiority of this method over that of making a cer- tain weight of solution (say 500 gr.) lies in the fact that measur- ing is less tedious than weighing. Problem. — ^A physician orders 120 c.c. of 2% solution of mercuric chloride. How can the order be filled? Calculation. — Let 120 c.c. represent 98%. Since I20 C.C. of water weigh 120 Gm., we have — 98% : 2% : : 120 Gm. : x (2.449 Gm.) Accordingly we dissolve 2.449 Gm. in 120 c.c. If pre- cisely 120 c.c. of product is to be dispensed, this volume may be measured out and the excess, which represent no commercial value, may be rejected. Note. — Percentage calculations of constituent are always based upon quantity of solution, if this is ascertainable in weight. So generally the known percentage will be 100%. But if the weight of solution (as in these problems) is an unknown, and Indeterminable quantity, the calculation is based upon weight of solvent. As the entire solution = 100%, the solvent = 100% — % of active constituent. The student should therefore have no difficulty in deciding when to use 100%, and when to use a lesser %. as the basis for his calculations. Problems. How much solvent, and how much constituents should be used to fill the following orders : 53. 20 C.C. of 5% solution of apomorphine hydrochlor. 54. 60 c.c. of 6% solution of cocaine hydrochloride. 55. I L. of 10% solution of camphor in cotton seed oil, the latter having a sp. gr. of .92. Ans. 102.22 Gm. 56. 2 f 5 5% solution of iodine in alcohol, sp. gr. of latter, .81. Solution.— 2 f S = 2 X 454.6 gr. X .81 = 73G.4 gr. 95% : 5% :: 736.4 gr. : (x) 38.7 gr. Accordingly 38.7 gr. of iodine are dissolved in 2 f 5 of alcohol. 57. 2 f 3 of 4% solution of homatropine. Ans. 4j4j gr. of homatropine and 2 f 5 of water. Weight to Volume Solutions 87 58. I f 5 of 20% solution of tannic acid in glycerin, sp. gr. of latter, 1.25 Ans. 142.38 gr. of tannic acid and I f 5 of glycerin. Weight to Volume Solutions. Percentage solutions, as has been stated, are construct- ed on the scale of 100 weight-units. As liquid medicines are administered by volume, both per mouth and hypoder- mically, the dose calculations become simplified if the strength of the liquid is expressed, not in per cent., but in Gm. in 100 c.c, or in gr. in 100 TTt. The strength of most galenical preparations is expressed in this way. And most physicians would prefer to have hypodermic solutions con- structed on this plan, so that by simple division the volume containing the desired dose of constituent may be calculated. In case of metric units weight to volume solutions dif- fer from the corresponding percentage solutions only as the sp. gr. of the solution differs from i.ooo. But in case of apothecaries' units there is a further difference, due to the fact that I nt of water weighs less than i gr., namely about .95 gr. In some localities much confusion has resulted from the misapplication of the conventional percentage symbol (%), this being used to express the strength of weight to volume solutions. In other words, some physicians use the % sym- bol, expecting the pharmacist to give it a special intei*- pretation. To avoid the resulting confusion the writer suggests that a new symbol, /^ , be adopted for weight to volume solutions. In case of metric prescriptions the w would stand for Gm., and the v for 100 c.c. ; while in case of apothecaries' units w would mean gr., and v, 100 Til. A 4 gr. to 100 IT, solution would accordingly be 4 /^ , and nothing would hi left to guess-work. Problems. 60. How many gr. of atropine are required to make I f 5 of a 4 /^ solution? 88 Pharmaceutical Arithmetic Then 100 tn, : 480 tq. :: 4 gr. : x (19^ gr.)* Or: If 100 Ttl of solution contain 4 gr. of constituent, [4 gr. -i- 100 =] .04 gr. is the amount of constituent in each TI^,; and .04 gr, X480 = 19.2 gr., the amount in 480 nv,. Problem!. 6i. How many g^. of cocaine hydrochloride are re- quired to make 4 f 3 of a 6 y^ solution ? Ans. 144 gr. 62. How many f 5 of a 4 ^^ solution could be made from I oz. of cocaine hydrochloride? Ans. 22fS6f3 17TIX. 63. What would be the /^ strength of a solution of boric add made by dissolving i oz. of the latter in enough water to make 2 pints of solution ? Ans. 2.84 /^ Other Methods for Expression of Strength of Solutions. To avoid fractions, the strength of very dilute soluticMis is sometimes expressed in parts in 1000, in 2000, 5000, or even in loooo. The solution of mercuric chloride, used as an antiseptic by surgeons, offers a case in point. As such solutions are generally for external use, the parts referred to are parts by weight. But the sp. gr. of these very dilute solutions is so nearly 1. 000, that I c.c. may be considered as equivalent to i Gm. in all cases, and I f 5 as equivalent to 454.6 gr. Problem!. 64. How many gr. of mercuric chloride are required to make i lb. of a i in 1000 solution? *Iii y^ solutions c. c of solution are proportional to Gm. of oonstitaeot; and at solution to sr. of constituent. Other Methods for Expression of Strength of Solutions 89 65. How many Gm. are required to make 2 L. of a i in 3000 solution? 66. How many gr. are required to make i pint of a i in 1000 solution ? Ans. y.2y gr. 67. How many gr. of mercuric chloride are required to make 2 f 5 of a solution, i f 3 of which diluted to ^ pint would make a I gr. to 2000 TH, solution? Solution. — Since % pint = 64 f 3, the solution should be 64 times as strong as a 1 gr. in 2000 Tli solution, that is, it should be a 64 gr. in 2000 in, solution. Then— 2000 Til : 960 HI : : 64 gr. : (x) 30.72 gr. 68. How many Gm. of mercuric chloride should be weighed out to make 120 c.c. of a solution 25 c.c. of which diluted to a L. would make a I Gm. in 3000 c.c. solution? Ans. 1.6 Gm. CHAPTER VII. CONCENTRATION AND DILUTION. If 100 Gm. of a io% salt solution is diluted to 200 Gm., that is, to double its original weight, the percentage of salt will thereby be reduced to 5%, that is, to one-half the orig- inal percentage. If the solution is diluted to four times its original weight, the percentage strength falls to one-fourth of the original percentage strength. If on the other hand the 10% salt solution is concentrated from 100 Gm. to 50 Gm. [the solvent being evaporated off without effecting loss of salt], the percentage strength is thereby doubled — ^be- comes 20%. Thus it is seen that with the amount of con- stituent remaining unchanged, the percentage strength de- creases in the same ratio in which the quantity of solution is increased, and increases in the same ratio in which the quantity of solution is decreased. As quantity of solution and percentage strength change in the same ratio, these two values are proportional to each other; but as the change is in opposite directions, they are said to be inversely proportional. However, the general rules for stating proportions, as given on page 6g are ap- plicable to inverse proportions as well as to direct propor- tions. With the amount of constituent (C) remaining con- stant, the following factors come into consideration in con- centration and dilution problems: 1. Initial amount of solution (or mixture, or prepara- tion) . 2. Amount after dilution or concentration. 3. Initial percentage strength of solution (or mixture, or preparation). 4. Percentage strength after dilution or concentration. If any three of these factors are given, the fourth may be found by proportion. Concentration and Dilution 91 If amount of diluent (to be added) is required, this is found by subtracting the initial amount of solution (or mix- ture, or preparation) from the amount after dilution. N. B. — If the diluent carries active constituent, as when a stronger opium is diluted with a weaker opium, these rules do not apply. For such problems see Chapter VIII. Since percentage is commonly weight percentage, the amounts must be in weight except when the percentage is expressly stated to be volume percentage, in which cases the amoimts must be In volume. Problems. 1. What is the % strength of a salt solution obtained bjy diluting 10 lb. of a 10% solution to 14 lb. ? Solution. — Percentage is asked for. Hence — : :: 10% : x After dilution the strength will be less than 10%. Hence the smaller of the remaining terms is the second, and the larger the first: 14 lb. : 10 lb. :: 10% : (x) 7|% 2. What is the % strength of a salt solution obtained by evaporating 10 lb. of a 10% solution to 8 lb. ? Solution. — Percentage is asked for. Hence — : :: lo/o : x After concentration the strength will be over 10%. Hence — 81b. : ID lb. :: 10% : (x) i2>^%. 3. To what weight must 500 Gm. of 10% salt solution be evaporated in order to make a 16% solution? Solution. — ^Weight is asked for ; and after concentration the solution will weigh less than 500 Gm. Hence — 16% : 10% :: 500 Gm. : (x) 312.5 Gm. 4. How many Gm. of 10% nitric acid (the diluted nitric acid of the U. S. P.) can be made from 300 Gm. of nitric acid containing 68% of abs. acid ? Ans. 2040 Gm. 5. How many lb. of acetic acid, 32% strong, are required to make 10 lb. of diluted acetic acid, which con- tains 6% of abs. acid? Ans. 1% lb. 6. What is the % strength of ammonia water obtained 92 Pharmaceutical Arithmetic by diluting 500 Gm. of stronger ammonia water, containing 28% of ammonia, to i8oo Gm.? Ans. 7^.%. 7. In order to bring a certain extract of opium to the official morphine strength (20%) i lb. 4 oz. had to be evaporated to 14 oz. What was the morphine strength of the extract prior to the evaporation? Ans. 14%. 8. To what weight must i Kg. of alcohol, 91%, be diluted to make alcohol of 41% strength?* 9. How many Gm. of sulphuric acid, 70%, can be made from 4 lb. of acid of 92.5% strength? N. B. — ^The two terms of a couplet must be in the same unit. 10. How many lb. of hydrochloric acid, 31.9%, are required to make 5 Kg. of 10% hydrochloric acid? 11. How many gallons of alcohol, 94% by vol., are re- quired to make 25 L. of alcohol which is 50% strong by vol.? Ans. 3.512 gal. 12. To what weight must 600 c.c. of solution of . so- dium hydroxide (sp. gr., 1.437; strength, 40%) be diluted in order to make a 5% solution? Ans. 6897.6 Gm. N. BL — ^Amount must b« ia weight for proportion. 13. What is the % strength of a diluted hydrochloric acid obtained by diluting i L. of hydrochloric acid (sp. g^., 1. 158; strength, 31.9%) to 5 lb.? Ans. 16.3%. 14. Diluted hydrocyanic acid should contain 2% of abs. acid. To how many Gm. must i lb. 5 oz. 200 gr. of a 2.35% acid be diluted in order to reduce it to official strength i. e., to 2% ? 15. To how many f 5 must 500 c.c. of alcohol (sp. gr., .82; strength, 91%) be diluted in order to make 58% alco- hol having a sp. gr. of .9? (See next page.) * Some text books give this rule for diluting alcohol : Designate the weight-percestage ot the stronger alcohol hj W, and the d& sired strength by v). Rule.— Mix v> parts bj weight of the stronger alcohol with water to make W parts by weight o( the product. To anyone familiar with the gaieral directions for stating proportions, thia rule is superfluous. Concentration and Dilution 93 Suggestion. — Convert 500 c.c. to Gm. ; find weight of dilution by proportion. Answer is in Gm. Convert this to c.c. ; then to f §• Ans. 24.17 f g. 16. A certain lump of opium on assaying is found to contain 9% of morphine, and 22% of moisture. What will be the morphine strength of the opium after drying [to make powdered opium] ? Solution. — ^If the opium contains 22 per cent, of moisture, 100 parts will make 78 parts of the dried opium. Since no morphine is lost in the drying, the percentage strength will be higher, i. e.^ the answer will be greater than 9 per cent. Hence — 78 parts : 100 parts : : 9% : (x) 11.53% 17. The official extract of nux vomica is a powdered extract, and is standardized to contain 5% of strychnine. On evaporating the natural extract to dryness, it usually yields a product containing more than 5% of strychnine. The di- luent used is milk sugfar; and to insure its thorough incor- poration, it is to be added while the extract is still of the consistency of a syrup. To calculate at this point the amount of milk sugar required, it is necessary to know the strychnine strength of the syrupy extract, and also the percentage of moisture present. Problem. — The syrupy extract of nux vomica is found to contain 4.5% of strychnine and 20% of moisture. How much milk sugar must be added to 600 Gm. of the extract in order to yield a product which after drying will contain 5% of strychnine? Solution. — (1). If the moisture amounts to 20 %, the yield of dry extract Is 80% of 600 Gm. 100% : 80% : : 600 Gm. : (x) 480 Gm. % of syr. ext. % of dry ext. wt. of eyr. ext. wt. of dry ext. (2). The strength of the dry extract is greater than 4.5%. Hence — 480 Gm. : 600 Gm. : 4.5 : (x) 5.625 (3) If the strength of this dry extract Is to be lowered from 6.625 per cent, to 5 per cent., the extract must be diluted to — 5% : 5.625% : : 480 Gm. : (x) 540 Gm. (4). To dilute the extract from 480 Gm. to 540 Gm., [540 Gm. ■*- 480 Gm. =] 60 Gm. of milk sugar must be added. 9fl PharmaceuHcal Arithmetic fThe four steps of the process are: (i). Finding the amount of dry extract the syrupy extract will yield. [(2.) Finding the % strengtK of the dry extract XsO. Finding the weight to which the dry extract would require dilution to reduce the strength to the required strength. [(4.) Finding by difference the amount of diluent to be added. 18. A certain lot of syrupy extract contains 12% of alkaloids and 46% of moisture. How much milk sugar must be added to 5 lb. 6 oz. to yield a product containing 15% of alkaloids after drying? Ans. 22.35 oz. 19. A certain lump of opium contains 28 % of mois- ture, and 8% of morphine, (a.) How much morphine could be obtained from 500 Gm. of dried opium prepared from this lump? b.) How much milk sugar would be required to reduce 200 Gm. of the dried opium to a morphine strength of 10% ? Ans. (a.) 55.55 Gm.; (b,) 22.22 Gm. 20. A certain fluid extract of scopola contains 20% of extractive, i. e., solid matter, and assays .5% of mydriatic alkaloids. How much milk sugar must be added to the res- idue obtained from i Kilo, of the fluid extract, in order to yield a solid extract assaying 2% of alkaloids ? Ans. 50 Gm. 21. To make 100 c.c. of paregoric of the official strength, 4 Gm. of opium containing not less than 10% of morphine, must be used. If the opium contains only 9% of morphine, how much should be used for each 100 c.c. of paregoric ? 22. A Liter of a certmn tincture is to contain 20 Gm. of drug, the standard alkaloid-strength of which is 1.5%. If the tincture were to be made from an extract (of the drug) of the alkaloid-content of 12%', how much of this extract should be used for each Liter of tincture? CHAPTER VIII. ALLIGATION.* Alligation Medial. Computing theJ Percentage Strength of a Mixture, or Solution, When the Amounts and the Percent- age Strengths op the Several Ingredients ARE Given. Problem. — ^A wholesale druggist mixes 5 lb. of opium containing 9% of morphine, 3 lb, of opium containing 6% of morphine, and 10 lb. of opium containing 12%. What is che morphine strength of the mixture? Solution. — 5 lb. of 9 per cent, opium contain as much mor- phine as [5 X 9 =] 45 lb. of 1 per cent, opium v.ould contain. 3 lb. of 6 per cent, opium contain as mr.ch morphine as [3 X 6 =] 18 lb. of 1 per cent, opium. 10 lb. of 12 per cent, opium contain as much morphine as [10 X 12 =] 120 lb. of 1 per cent, opium. Then the 18 lb. of product [5 lb. + 3 lb. + 10 lb.] contain as much morphine as 183 lb. [45 lb. + 18 lb. + 120 lb.] of 1 per cent, opiui would contain. And the morphine strength of the product must be as many times 1 per cent, as 18 lb. is contained in 183 lb. 183 -=- 18 = 10.16, Hence the morphine strength of the product is 10.16 per cent. This operation may be made the basis of the following rule: I. Multiply the amounts of the several ingredients (all of which must be expressed in the same unit) by their per- centage strengths. •From the Latin alUgare, to bind: — ^referring to the linking togfether o( ▼alues in alligation alternate, as shown on page 90. g6 Pharmaceutical Arithmetic 5[lb.] X 9[%] = 45- [% units]. 3[lb.] X 6[%] = i8. [% units]. 10 [lb.] X I2[%] = 120. [% units]. 2. Add the products. 45 -}- i8 + 120 = 183 [% units], 3. Add also the amounts. 5 + 3 + 10 = i8[lb.] 4. Divide the sum of the products by the sum of the amounts. 183 -f- 18 = 10.16. The quotient is the percentage strength of the mixture. Note. — ^In case the strengths are given in Tolume-percentage, the amounts must be in volume; if in weight- percentage, th« amounts must be in weight, as in the example. Problems. 1. A drug house receives two shipments of cinchona. The first, consisting of 500 lb., contains on an average 3.3% of quinine. The second, consisting of 300 lb., contains 2.1% of quinine. What is the quinine strength of the product ob- tained by mixing the two lots? Ans. 2.85%. 2. A druggist finds in his laboratory the following lots of diluted alcohol: 1500 cc, 60% by vol. strong, 400 cc, 40% by vol. strong, i pint, 50% by vol. strong. What is the strength of the product obtained by mixing the three lots? Ans. 54.63%. Computing the Specific Gravity of a Mixture When THE Amounts and the Specific Gravities of the Ingredients are Given. The principles of alligation medial apply also to spedfic gravities. But it is required that the amounts be in volume. Alligation Medial 97 and not in weight ; for specific gravity expresses the weight of a unit of volume — not of a unit of weight.* 3. What would be the sp. gr. of the product obtained by mixing 3000 c. c. of benzin and 1200 c.c. of naphtha, the former having a sp. gr. of .67, and the latter of .718? Solution— 3000 c.c. X .67 [sp. gr.] = 2010 Gm. 1200 c.c X .718 [sp. gr.] = 861.6 Gm. 4200 c. c 2871.6 Gm. 2871.6 -=- 4200 = .683 (sp. gr. of mixture). 4. What would be the sp. gr. of a mixture of 2 f § of bromoform, sp. gr., 2.9, and 8 f § of chloroform, sp. gr., 1.48? Ans. 1.764. 5. What would be the sp. gr. of a mixture of 150 Gm. of ether, sp. gr., .716 and 200 Gm. of chloroform, sp. gr., 1.48? Suggestion. — Reduce Gm. to c.c; then proceed by al- ligation medial. Ans. 1.015. 6. What would be the sp. gr. of a mixture of i pint of alcohol, sp. gr., .81, and i lb. of ether, sp. gr., .716? Ans. .756. •A cubic centimeter of water weighs 1 Gramme. A c.c. of a liquid having a Bp. gr. of 3 , weighs 3 Gm. On mixing 1 c.c. of water, sp. gr. 1, with 1 c.c. of the liquid, sp. gr., 8, we get 2 c.c. of product, weighing 1 + 3=4 Qm. II 2 c.c. weigh 4 Gm., 1 c.c. will weigh 4 Gm. -r- 2 = 2 Gm. Then the sp. gr. of the product must be 2, which by laboratory experiment can be proven to be correct. But suppose that 1 Gm. of water and 1 Gm. of the liquid are miJted. What will be the sp. gr. of the product? Misapplying the rxile to weights we obtain the same answer as before — name- ly, sp. gr., 2. This cannot be correct if the answer to the first problem is correct; for the proportion of 1 Gm. of water to 1 Gm. of the liquid is the same as 1 c.c. of water to . }^ c.c. of the liquid, while in the first problem we have 1 c.c. of each. But oh reducing the weights to volumes, and then applying the rule, we obtain for an answer the sp. gr, of 1.6. 1 c.c. X 1 tsp. gr.] = 1 Gm. }< CO. X 3 [sp. gr.] = 1 Gm. Then 2 -r- 1 ^ = 1.5. [sp. gr. of mixt.] 1 ^ C.C. 2 Gm. This by laboratory experiment can be proven to be correct. [In practice It ia difficult to find two liquids of which one has a sp. gr. just three times that of the other. Even numbers were used in the example for sake of brevity and simplicity. For the laboratory experiment any two liquids which are miscible. and differ considerably in sp. g:r., may be used.] 98 Pharmaceutical Arithmetic Alligation Alternate. To Find the Proportion in Which Ingredients o^ Known Strength Must be Mixed to make A Mixture op Required Strength. Problem. — In what proportion must opium lo% strong and opium 14% strong be mixed to make opium 13% strong ?* Solution. — ^The strength of the stronger opium is 1% too high; that of the weaker opium is 3% too low. Since the difference between the strength of the stronger opium and the required strength is just ^ as great as the difference between the strength of the weaker opium and the required strength, three parts of the stronger opium must be taken to one part of the weaker. In other words, the excess in strength of three parts of the stronger opium will just bal- ance the deficiency of one part of the weaker opium. Problem. — In what proportion must opium 10% strong and opium 15% strong be mixed to make opium 13% strong ? Solution. — The weaker opium is 3% too weak: the stronger is 2% too strong. Then the excess in strength of 3 parts of the stronger will just balance the deficiency of 2 parts of the weaker. For 2 [parts] X 3 [%] =6; and 3 [parts] X 2 [%] = 6. [The same principle applies in alligation alternate as is applied in writing chemical formulae. — The formula for zinc phosphide is Zug Pj. Now zinc has 2 valencies, and phosphorus has 3. Hence 3 atoms of bivalent zinc have as many valencies (6) as 2 atoms of trivalent phosphorus. In like manner 3 parts of the opium which is 2% too strong will fortify to the required strength 2 parts of the opium which is 3% too weak.] Generai, Rule. Per cent, of stronger ingredient minus desired per cent, equals parts of weaker ingredient. Per cent, desired minus per cent, of weaker ingredient equals parts of stronger ingredient. *0fficia] opium contains from 12 to 12.6^ of morphine. 'Alligation ^Alternate 99 In case there are more than two ingredients, their per- centage strengths must be paired off so that the difference between each percentage strength and the required per- centage strength is just compensated for by the proper number of parts of the opposite kind. Problem. — In what proportion must alcohol, 20%- strong, alcohol, 40% strong, alcohol, 65% strong, and al- cohol, 90% strong, be mixed to make alcohol which is 50%^ strong? Proceed as follows: I. Write the percentages in a column [preferably in numerical order]. Place a brace [-{ ] in front of the col- umn, and in front of the brace write the required percentage. Thus— 50^ 2. Connect with a line each percentage which is great- er than the required percentage with one that is less than the required percentage; and each one that is less with one that is greater than the required percentage. Thus — f r-20% f p20^ 1^ L.90% I . i%% 3. Write the difference between the first percentage and the required percentage opposite the percentage connect- ed with the first percentage by a line. Proceed in this manner with all the percentages, in each case writing the difference not opposite the percentage compared, but op- posite the percentage at the other end of the line. Thus — ■ r-20%= 40 .40%= 15 ^5%= 10 ^90^=30 lOO Pharmaceutical Arithmetic The difference between 20% and 50% being 30%, 30 is placed opposite 90%. The difference between 40% and 50% being 10%, 10 is placed opposite the 65%. The difference between 65% and 50% being 15, 15 is placed opposite the 40%. The difference between 90% and 50% being 40, 40 is placed opposite the 20%. The other answer is— -20 = 15 '•90 = 10 The differences denote the number of parts which should be used of the several ingredients to make a mixture of the desired strength. In case of weight-percentage the proportionate parts are parts by weight; and in case of volume percentage they are parts by volume. Proof. For first answer. By Alligation Medial. 40 [parts] X 20 [%] = 800 [% units] 15 [parts] X 40 [%] = 600 [% units] 10 [parts] X 65 [%] = 650 [% units] 30 [parts] X 90 [%] = 2700 [% units] 95 [parts] contain 4750 [% units] Then— 4750 ^ 95 = 50%. Remarks. — ^When there are more than three ingredients, a number of correct answers are possible, because the per- centages may be paired off in different ways. And even in case of but three ingredients — when but one answer is ob- tainable by alligation — the ingredients may be mixed in an indefinite number of proportions. So the answer obtained by alligation, in case of three ingredients, is not the only cor- rect answer: it is a correct answer. And the answers ob- tained by alligation in case of four or more ingredients, are not all, but only some, of the correct answers. Alligation Alternate lOI In Case Either the Weaker or the Stronger Ingredi- ents Exceed in Number. Problem. — ^A wholesale druggist has five lots of opium— 8%, 10%, 11%, 14% and 15% morphine strength respec- tively. In what proportion may these be mixed in order to make a product of 12% morphine strength? Remember that in no case can a weaker % be connected with a weaker, or a stronger with a stronger. Proceed as in the preceding example, except that ooe of the percentages must be connected with two others : 12% 8% = 3 —10% = 2 rll% =2 tU% =2 + 1 -Ibfc == 4 As the 14% is connected with two other percentages, the difference between 14% and the required percentage must be placed opposite both the percentages thus connected ; i. e., opposite the 10%, and the 11%. And the difference be- tween 10% and 12%, and the difference between the li% and 12%, are both to be placed opposite the 14%. Question.— 'iHovf many correct answers to this problem are obtainable by alligation alternate? Problems. 7. In what proportion should alcohol, 91% strong, be mixed with alcohol, 48% strong, to make alcohol 70% strong? Ans. 22 parts of 91% and 21 parts of 48%. 8. In what proportion must cinchona containing 11% of total alkaloid, cinchona containing 8%, and cinchona con- taining 4%, be mixed to make a product containing 5% of alkaloids? Ans. i part of 11%, I part of 8%, and 9 parts of 4%. 9. In what proportion should jalap containing 5% resin, jalap containing 7%, jalap containing 4%, and jalap 102 Pharmaceutical Arithmetic containing 3%, be mixed to make a product containing 6% ? In case of fractional parts. 10. In what proportion must coca .8% strong be mixed with coca .38% strong to make a product .5% strong? Solution.— f.38 per cent. = .30 parts, which X 100 = 30 parts. .80 per cent. = .12 parts, which X 100 = 12 parts. Note. — ^As fractions are cumbersome, eliminate them by multi- plication. In this case multiply by 100. 11. In what proportion should belladonna .38%, bella- donna, .18%, and belladonna, .30%, be mixed to make a product .35% strong? Ans. 22 parts of .38%, 3 parts of each of the others. In Cask Active Constituents, or Diluents, or Both ARE Ingredients in the Mixture. Problem. — In what proportion must chloroform be added to a 20% solution of chloroform to increase the strength of the latter to 35%? Solution. — ^The chloroform figures in the calculation as a 100 per cent, solution, thus — [ 20 per cent. = 66 parts = 13 parts. 36 per cent. < or by cancellation ^ 100 per cent. =: 15 parts = 3 parts. TTien 65 parts of solution require 15 parts of chloroform. Or, reducing the ratio to its simplest expression by cancellation — 13 parts of solution require 3 parts of chloroform. Problem — In what proportion should alcohol, 91% strong, alcohol, 48% strong, and water be mixed to maike a product 35% strong? 'Alligation Alternate 103 Solution. — ^The water maj be considered as an alcohol of zero percentage. Then — 35^ f=0% (water) « 13 + 56 = 69 parts. L48%=:"" 48fo —35 part*. L.9lfo = 35 parts. Problems. 12. In what proportion must potassium iodide be added to a potassium iodide solution 15% strong in order to in- crease its strength to 25% ? Ans. 2 parts of potass, iodide to 15 parts of 15% solution. 13. In what proportion should opium 10% strong, opium 15% strong, and milk sugar be mixed to make a pro- duct 13% strong? Ans. 2 parts of 10%, 16 parts of 15%, and 2 parts of milk sugar. 14. In what proportion should absolute acetic acid, acetic acid 36% strong, and water be mixed to make acetic acid 60% strong? To Find ths Proportion in Which Ingredients op Giv- en Specific Gravities Must be Mixed to Makb A Mixture of Given Specific Gravity. Alligation alternate is applicable to volumes of liquids and their specific gravities, provided a change in volume, due to chemical action, does not take place.* Specific gravities of mixtures of alcohol and water can- not be calculated. See U. S. P. alcohol table. Problem. — In what proportion by volume must petro- latum, sp. gr., .82, be mixed with petrolatum, sp. gj., .85, to make a product having the sp. gr. of .83 ? Solution. — J , .85 = .01 X 100 = 1. parta. (r.85 = .C *• \ L.82 = .02 X 100 = 2. parts. Eliminating the fractions, the answer is, 2 parts by *The parts calculated from ipeciflo gniTitics are in every case parts bf Tolume. See page 97. 104 Pharmaceutical Arithmetic volume of the lighter to i part by volume of the heavier pa- trolatmn. Problems. 15. A mixture of ether (.716) and absolute alcohol (.79) has a sp. gr. of .75. What proportion by volume of ether does it contain ? Ans. 40 vol. of ether to 34 vol. of abs. al- cohol. 16. In what proportion by volume should glycerin, sp. g^., 1.25, alcohol, sp. g^., .82, be mixed to make a product having a sp. gr. of .95? Ans. Glycerin 13 p., alcohol, 30 p. 17. In stock we have two solutions of soda, the sp. g^. of one being 1.06, tliat of the other, 1.5. In what proportion must these be mixed to yield a product of the sp. gr. of 1.225? Ans. 275 vol. of 1.06 to 165 vol. 1.5, which reduced to simplest expression gives 5 vol. to 3 vol. In Case the Quantity of One op the Ingredients is Specified, and the Quantity of the Other Ingredients is to be Cai^culated. Problem. — How much scammony having a resin-con- tent of 82% must be mixed with 5 lb. of scammony having a resin-content of 93%, to make a product having a resin- content of 90% ? Solution. — By alligation alternate we determine in what proportion the two qualities must be mixed : 90 per cent. \ ^^ P«^ *=«^*- = ^ ?*"*«• I 93 per cent. = 8 parts. From the proportionate parts we calculate the quantities by proportion. The three known terms are — 3 parts, 8 parts, and 5 lb. Since quantity (weight) is asked for, 5 lb. must be the third term. The quantity of the weaker scammony is asked for ; and as this is to be used in the proportion of 3 parts to 8 of the other, 3 being less than 8, the answer is to be smaller than 5 lb. Hence — 8 parts : 3 parts : : 5 lb. : (x) 1% lb. = 1 lb. 14 oz. In order that confusion may be avoided it is well to place the given quantity opposite the proportionate parts Alligation Alternate 10$ which are to be taken of the ingredient for which the quan- tity is given, and to place an interrogation point to indicate the quantities sought; thus — ^ ( 82% = 3 parts = ? ^ ) 93% = 8 parts = 5 lb. The known terms may now be located without difficul- ty; and whether the answer should be greater or smaller than the given quantity — the third term — may be seen by inspection. The interrogation point follows "3 parts" in the example; a given quantity follows "8 parts." Since 3 is smaller than 8, the answer must be smaller than 5 lb.— the given quantity. Problem — ^How much alcohol, 45% strong, how much 60% strong, and how much water, should be mixed with 500 Gm. of alcohol, 90% strong, in order that the product may have a strength of 50% ? Solution. — ( r— 0% (water) = 40 parts = t «in«, J r45% = 10 parts = ? ^"^1 L60% = 6 parts = T [ I — 90% = 60 parts = 500 Gm Then by proportion*: 50 parts : 30 parts 60 parts ; 10 parts 60 parts : 6 parts 500 Gm. : x (400 Gm.) 500 Gm. : x (100 Gm.) 500 Gm. : X ( 50 Gm.) and Proof. 400 Gm. X 0% =: 0% unit 100 Gm. X 45% = 4500% units 50 Gm. X 60% =: 3000% units 500 Gm. X 90% = 45000% units 1050 Gm. = 52500% units 52500 -f- 1050 = 50 [%]. • The general proportion is — parts of ingredient, quantity of which is given : parts of ingredientt . quantity of which is sought : : quantity given : x z = quantity sought. io6 Pharmaceutical Arithmetic Problems. 1 8. How many Gm. of nitric acid, 68% strong, must be added to 500 Gm. of acid, 10% strong, to make a pro- duct 50% strong? Ans. iiii.iii Gm. 19. How many Gm. of sulphuric add, 90% strong, must be added to I L. of water to make an acid 10% strong ? 20. How many c.c of water should be added to i L. of a solution of soda, sp. gr., 14, in order to make a solution of the sp. gr. of 1.2? 21. How many f^ of ammonia water, sp. gr., .96, must be added to 8 f^ of ammonia water, sp. gr., .92, to make a product having a sp. gr. of .95? Ans. 24. f^. 22. How many c.c. of ether, sp. gr., .716, must be added to i L. of chloroform, sp. gr., 1.48, to yield a product of the sp. gr. of 1.25? Ans. 430.7 cc In Case the Quantity is Specified for Severai, Ingredients. Problem. — ^We have 5 lb. of acetic add, 28% strong, 3 lb. of acetic acid, 20% strong, and i lb., 10% strong. How much gladal acetic add, 99% strong, should be added to a mixture of these in order that the product may have the strength of 36% ? Solution.— 5 lb. X 28 % = 140 % units 3 lb. X 20 % = 60 % units. 1 lb. X 10 % = 10 % units. 9 lb. = 210% units 210 -5- 9 = 23 » % A mixture of the given quantities would be 23^^% strong, and the weight of the mixtures would be 9 lb. The problem has now been simplified to — How much acetic acid 99% strong must be added to 9 lb. of add 23^% strong, to make a product 36% strong? Alligation Alternate I02. -, -, 5 23J% — 63 parts — 9 ^>J99 ^ =- 12f parts « ? lb. Then— 63 parts : 12} parts : : 9 lb. : (z) If } Ik =tl lb. 12 oz. 417 gr. Problems. 23. A druggist has 400 Gm. of opium, 10% strong, 350 Gm., 12% strong. How much opium, 16% strong, should he add to a mixture of the 10% and the 12% opium to make a product 13% strong? Ans. 516.66 Gm. 24. How much ammonia water, 28%, should be added to a mixture of 5 lb. of ammonia water, 10%, and 300 Gm., 8%, to make a product 15% strong? 25. In a laboratory there are three lots of solution of potassium iodide as follows: — ^400 Gm. of 5%, 200 Gm. of 12%, and 300 Gm. of 15%. How much potassium iodide must be dissolved in a mixture of the three lots to make a product 25% strong? Ans. 181.33 Gm. In Case the Quantity op Product Wantej> is Specified, AND Quantity of Each of the Ingredients is to BE Calculated. Problem. — How much cinchona, quinine-strength, 3.9% and how much cinchona, 2.4%, must be used to make 10 lb. of a product having a quinine-strength of 3% ? SolutIoii.-3% f -^^^ = .6 X 10 = 6 parts = ? '*' }2A7o = .9 X 10 =_9^ parts = ? 15 parts = 10 lb. If 6 parts and 9 parts are mixed, the product will be 15 parts. Then to make 15 parts of the product 9 parts of the cinchona, 2.4%, must be used; and from these 9 parts the quantity for 10 lb. may be found by proportion, weights being proportional to proportionate parts. io8 Pharmaceutical Arithmetic Thu&— 15 parts : 9 parts :: 10 lb. : (x) 6 lb* In like manner the quantity of the stronger cinchon calculated: 15 parts : 6 parts : : 10 lb. : (x) 4 lb • Proof. 6 lb. X 2-4% = 14.4% units. 4 lb. X 3-9% = 15-6% units. 10 lb. 30.0% units. 30 -7- 10 = 3% and 61b. + 4 lb. = 10 Ih. Problems. 26. How many Gm. of colchicum, alkaloidal strength, .4%, and how many Gm. of the strengfth of .56%, must be used to make 20 Kgm. of product of an alkaloidal strength of .5% ? Ans. 7500 Gm. of .4%, and 12500 Gm., of .56%. 27. How many gr. of mercuric chloride must be added to how many gr. of mercuric chloride solution, 2% strong, to make 500 gr. of a solution which is 5% strong? 28. 5 lb. of opium containing 14% of morphine is wanted. How much 9% opium, how much 11%, and how much 15.4% may be used to make a mixture of required weight and strength. Ans. of 9% and 1 1 % each .648 -}- lb. ; of 15.4 %, 37 + lb. —It will be obserred that the general proportion ia: Sum of the parts : parts of the incredient : : weight of the mixture : X. z = weight of the ingredient. Observe that the sum of the proportionate parts enters Into the calculation onlj when the quantity of the entire mixture is used as a basis for the calcula- tion of the amounts of the ingredients. When the amount of one ingredient if calculated from the amount of another ingredient, the sum is not used. 'Alligation Alternate 109 In Case Quantity of Product Wanted, and also the Quantity of One or More Ingredients is Specified. Problem. — 1000 Gm. of alcohol, 50% strong, is to be made from 100 Gm. of 80% al^%. Assuming the atomic weight of hydrogen to be I, carbon 12, and oxygen 16, what is the formula of the compound? Process. — First calculate how many parts of each of the three elements would be present in 90 parts, 90 being the molecular weight: 100% : 40 % :: 90 parts : x (36 parts). 100% : 6f % :: 90 parts : x ( 6 parts). 100% : 53>^% :: 90 parts : x (48 parts). Then divide the parts for each of the elements by the atomic weight of that element Thus — Carbon, 36 parts —- 12 (at wt.) = 3. Hydrogen, 6 parts -i- i (at wt.) = 6. Oxygen, 48 parts -r- 16 (at. wt.) = 3. The molecular formula is CaHaOj. Note. — ^The calculation is here simplified by use of approxi- mate atomio weights. In practice the accurate atomic weights are employed. Calculation op Molecular Weight prom Molecular Formula. As has been stated, the molecular weight of a compound must be determined experimentally before it becomes pos- sible to deduce the molecular formula. But when the latter Calculation of Molecular Weight 113 is known, the molecular weight may be calculated from the molecular formula ; for the molecular weight is the sum of the weights of the atoms in the molecule. For instance: The molecular weight of NaOH is the atomic weight of Na = 23, plus the atomic weight of 0=16, plus the atomic weight of H= 1.008, which gives a total of 40.008, or 40 if the approximate atomic weight of I is used for H. Problem. — Calculate the molecular weight of H2SO4. Solution. — H^= 1X2= 2 S =32 X 1 = 32 0, = 16X4 = 64 Mol. wt. of H2S04 = 98 Note. — The numbers in italics are the atomic weights. Problem. — Calculate the molecular weight of NaaCOg.ioHjO. Solution. — Na^ = 23 X 2 = 46 C = i2X 1 = 12 O3 = i6X 3 = 48 loHj^Hjo = 1X20 = 20 loO = i6X 10 = 160 . wt. of NajCOa.ioH^O^ 286 Mol. Problem. — Calculate the molecular weight of Mg(C03),.Mg(OH),.5HA Solution. — Reducing the rational formula to an empiri- cal formula gives Mg5C40i9Hi2. Mg^ = 24.32 X 5=121.60 C^ =12 X 4= 48 O18 =16 X 19=304 Hj2 = -T X 12= 12 Mol. wt. of (MgC03)4.Mg(OH)2.5H20 = 485.6o 114 Pharmaceutical Arithmetic Problems. These answers were obtained by use of accurate atomic weights. The student may, however, shorten the work of figur- ing' by using approximate atomic weights in all cases when the calculations are performed merely for practice, and not in con- nection with laboratory exercises. 1. Calculate the molecular weight for (a.) HjO; (b.) NaOH; (c.) HgC^; (d.) QoH^. Ans. (d.) 128. 2. For— (a.) FeS0,.7H,0; (b.) Fe,{PU,0,),', (c.) HC,H„03. (d.) Al,K2(SOJ,.24H20. Ans. (d.) 949- 3. For— (a.) Q2H„0,i;(b.) CHCl,. Ans. (a.) 342. 4. For — C20H24N2O2.3H2O (Quinine). Ans. 378. 5. For— (C2xH22Na02)3H2S04.5HaO (Strychnine Sulphate). Ans. 856.5. Note. — ^Remember that a subscript following a symbol applies to that symbol only ; that a subscript following parentheses mul- tiplies all within the parentheses ; and that a coeflBcient preceding a formula multiplies the entire formula. If, however, the coeffi- cient precedes only part of a formula, only the portion Thich follows is affected by the coefficient. Thus in the formula K«Fe(CN),3HjO the subscript 4 multiplies K; the subscript 6 multiplies C and N ; and the coefficient 3 multiplies HjO. Calculation of the Percentage Composition of a Com- pound FROM ITS Formula. Problem. — Calculate the percentage composition of FejOa. Solution. — Fcj = 35.8 X 2 = 1 1 1 .6 03=16 X3= 48 Mol. wt. of Fe203 = i59.6 Atomic weights, and hence also molecular weights, may be considered as parts by weight. It follows then that 159.6 parts of FcaOj contains 11 1.6 parts of Fe. From these data the percentage of Fe may be calculated by proportion in the usual manner : The three known terms are, 159.6 parts, 11 1.6 parts, and 100%. The answer is to express %. Hence — : :: 100% : x Calculation of the Percentage Composition 115 The answer is to be smaller than 100%, because a part must be smaller than the whole. Hence — 159.6 parts : 11 1.6 parts :: 100% : x X = 69.9% = % of Fe. In like manner the percentage of O may be calculated : 159.6 parts : 48 parts :: 100% : x X = 30.07% = % of O. Problem. — ^What is the percentage composition of Solution. — Ca=J^X 6= 72 Hi2= iX 12= 12 CeH Oe =i(5X 6 = = 96 180 Then— . 180 parts 180 parts 180 parts : 72 parts : 12 parts : 96 parts Problems. :: 100% : x x=:4o% = % of C. :: 100% : x x = 6.6% = % of H. :: 100% : x x=53-3% = % oiO. 6. Calculate H,0; (b.) H3O2 the percentage ; (c.) H3PO,; [quinine]. Ans composition of — (a.) (d.) QA,0,,; (e.) (a.) ii.i + % of H and 88.8 + % of O. The Percentage-Content of Only One Element or Group of Elements is Required. Problem. — What is the percentage of iron in FeS04.7H20? Solution. — Fe= 35.84X1 = 55-84 S =3^-7 XI =32 0, = i6 X4 =64 7H2= I X 7X2= 14 7O =16 X 7 =112 Mol. wt. of FeS04.7H20 = 277.84 ii6 Pharmaceutical Arithmetic Then — 277.84 parts : 55.84 :: 100% : x x=:2i% = % of Fe. Problem. — Calculate the percentage of water of crystal- lization in NaaCOg.ioHaO. Solution. — The molecular weight of NaaCOg.ioHaO is 286.16. The molecular weight of H2O is 18; hence the relative weight for the ten molecules is 180. In other words, 286.16 parts of NajCOg.ioHjO con- tain 180 parts of water. Then — 286.16 parts : 180 parts :: 100% : x(62.9%). Problems. 7. What % of anhydrous FeClg in FeCl3.6H20? 8. What % of Fe in FeCl3.6H20? Ans. 20.69%. 9. What % of Cin CH,? 10. What % of I in CHI3 [iodoform] ? Ans. 96.6 4- %. 11. What % of I in QNHI^ [iodol] ? Ans. 8970. 12. What % of arsenic in realgar — an arsenical ore of the composition AsgSg ? 13. What % of zinc in sphalerite — a zinc ore of the composition of ZnS? Ans. 67%. 14. What % of water of crystallization in FeS04.7H20 ? 15. In making dried sodium carbonate the crystallized sodium carbonate is heated until it has lost one-half its weight. What % of water of crystallization remains in the dried carbonate ? Solution. — Mol. wt. of NaaCOa.ioHgO^: 286.16. If the weight is reduced to one-half, the molecular weight of the product must be 286.16-^2 = 143. The molecular weight of anhydrous Na2C03=io6. Then 143 minus 106 =37 = weight of water in the molecule of dried carbonate. And 143 parts : 37 parts :: 100% : x (25.87%). 16. How many molecules of water of crystallization in the dried sodium carbonate? Solution. — If the molecular weight is one-half that of the crystallized carbonate, there are about 37 parts of water present; for 143 — 106=37. An^ 37-^^^ (mol. wt. of Calculations Based on Chemical Reactions 117 H20)=2+. So a little over two molecules are present; and the formula Na2C03.2H20 is nearly correct for the; dried carbonate, U. S. P., 1890. Calculations Based on Chemical Reactions. Atomic weights are relative weights, and for arithmeti- cal purposes may be thought of as parts by weight. This being true of atomic weights, it must be true also for molec- ular weights and for multiples of molecular weights. By the equation — Zn-f-H2S04 = ZnS04-|-H2 it is indicated that an atom of zinc requires a molecule of absolute sulphuric acid in order to make a molecule of zinc sulphate, and two atoms [a molecule] of hydrogen.* Now if the atomic weight of Zn is 65.37, ^^^ the molecular weight of H2SO4 is 98, it follows that 65.37 parts by weight of Zn will require 98 parts by weight of H2SO4. And if the molecular weight of ZnS04 is 161.44, and the atomic weight of H is i, 161.44 parts by weight of ZnS04 and 2 parts by weight of H will be the products of the reaction. Thus — Zn -f H2SO4 = ZnSO^ -f H2. 65.37 +98 =161.44 + 2. Problem. — How many Gm. of abs. H2SO4 is required to convert 50 Gm. of Zn into ZnS04 ? Solution. — Since atomic and molecular weights may be considered as indicating parts by weight, the problem may be stated as follows : — if 65.37 parts of Zn require 98 parts of H2SO4, 50 Gm. of Zn will require how many Gm. of H2SO4 ? The parts, being parts by weight, are proportional to Gm., and the ans\Yer may be found by proportion in the usual manner, as follows : The answer is to express weight [Gm.] ; hence the known term expressing weight must be the third. : :: 50 Gm. : x • In order that the reaction may occur as shown in the equation it is necessary that the acid used be diluted with considerable water, concen- trated acid giving rise to other products, including SO2 and H2S. However, the water, though necessary, is the same on the product-side of the equa- tion as on the factor-side, and hence does not enter Into the calculations. ii8 Pharmaceutical Arithmetic The answer is to be larger than 50 Gm., because the answer is to give the amount of H2SO4, and as it requires 98 parts of acid for 65.37 parts of Zn ; i. e., it requires more acid than Zn, the answer will be larger than 50 Gm. Hence — 65.37 parts • 98 parts :: 50 Gm. : x(75Gm.). Problem. — How many Gm. of Zn will 75 Gm. of abso- lute H2SO4 convert into ZnS04 ? Solution. — ^Weight is asked for ; hence the known term expressing weight must be the third. The answer is to be smaller than 75 Gm., because it requires less Zn than H2SO1. Hence — 98 parts : 65.37 parts :: 75 Gm. : x (50 Gm.). In like manner — by proportion — ^the amount of ZnS04 formed may be calculated, as may also the amount of H evolved. Again, the amount of one of the products might be given. From it the amount of the other products may be calculated ; also the amount of Zn, and the amount of HjSO^ required. In short, */ the quantity for any one member of an equation is given, the quantity for any other member may be found by proportion. ' ' It will be seen by close inspection that twelve distinct problems may be based on the equation given ; and in case of equations of more than four members, the number of possible problems is correspondingly larger. Yet any and all may be solved by proportion, no specific rules being nec- essary. However, in order that there may be no mistake in selecting the terms for the proportion, the following pro- cedure is recommended: 1. Write the equation for the reaction on which the problem is to be based. 2. Select the two members of the equation which fig- ure in the problem ; namely, (i.) the compound [or element] the quantity of which is given in the problem, and (2.) the compound [or element] the quantity of which is sought Draw a line under each of the members of the equation thus selected; also indicate for which one the quantity is given and for which one it is sought 3. Calculate the reacting weight for each of these two Calculations Based on Chemical Reactions 119 members of the equation, writing the reacting weight under the lines. The term reacting weight Is here used to Indicate the atomic weight, molecular weight, or multiple of these, as required by the equation. If, in the latter, the molecule is multiplied by a coefficient, the mciecular weight must be multiplied by the same coefficient, anti the product used in the proportion. 4. Select the third term. When quantity is asked for, the quantity given in the problem must constitute the third term. However, since the other values are weights — react- ing weights, considered as carts by weight — ^the quantities must be in weight ; for volumes are not proportional to parts by weight. 5. Select the second term. Reacting weights [parts by weight] are directly proportional. Hence the answer will be larger than the third term if the reacting weight of the substance, the quantity of which is sought, exceeds the re- acting weight of the substance, the quantity of which is given. Or, in other words, the reacting weight of the sub- stance, the quantity of which is sought, must be the second term. 6. Then the reacting weight of the substance, the quantity of which is given, must be the first term, and the general proportion will be as follows: Reacting weight of substance quantity of which is given is to the reacting weight of the substance quantity of which is sought as the quantity given is to the quantity sought. This must be so, because the two values are directly proportional. Appucation op Ruia Problem. — ^How many Gm. of Zn are required to gen- erate 30 Gm. of H ? Solution. — 1. Equation: Zn -f H^SO* = H, + ZnSO^ 2. Members required: Zn -f- H2SO4 = Hj -j- ZnSO* sought given 120 Pharmaceutical Arithmetic 3- Reacting weights J Zn + H.SO. = H, + ZnSO. required: Y^^^ht g'ven 4. Third term selected : : : : 30 Gm. : x. 5. Second term selected : To generate 2 parts of H requires more than 2 parts of Zn ; therefore the answer must be larger than the third term. Then— 2 parts : 65.37 parts : : 30 Gm. : x (960.55 Gm.) That is — React, wt. H. : React. wt.Zn : : wt. H : wt. Zn. I J It will be seen that the IT and Zn alternate. This al- ternation of substances in the proportion — ^made necessary by the fact that the two values are directly proportional — may serve as proof that the first and second terms have been correctly placed. Problem. — How many Gm. of ferric sulphate are re- quired to make 100 Gm. of ferric hydroxide? Solution. — Equation : Fe2(S04)8+6NH40H=2Fe(OH)3+3(NH,)2SO,. sought given 399.9 213.7 Then 213.7 parts : 399.9 parts :: 100 Gm. : x (187 Gm.). R. wt. Fe(OH)s : R.wt,Fej(S0t)s : : Wt.Fe(OH), : Wt. Fe2(S04)s. Notice that the reacting weight of Fe(OH)3 is twice its molecular weight [106.86X2 = 213.7]; the equation showing that one molecule of FcjC 804)3 will make two molecules of Fe(OH)3. Note. — On comparing the amounts given in the working formulas of text-books with the amounts of the factors as cal- culated from equations, great discrepancies will often be noticed. Thus, the amount of iron used in making Felj is greatly in excess of the amount required by the equation. In like manner, am- monia water, when used to precipitate ferric salts, is used in Problems Based on Chemical Reactions xai considerable excess. Practical experience has proven such ex- cess necessary. [For explanations the student is referred to the standard works on theoretical chemistry, especially to the chapters on mass action, on the ion theory, and on chemical equilibrium.] How- ever, even in these instances the quantities as calculated from the equation, form a basis for the working formula. Problems. Some of the answers which follow were obtained with approxi- mate atomic weights. 17. How much lime(CaO) can be made from 100 Gm. of marble (CaCOa) ? Ans. 56.4 Gm. Equation: CaCOg + heat = CaO + CO,. 18. How much carbon dioxide (COj) is formed at the same time? 19. How much absolute HjSO^ is required to neutral- ize 50 Gm. of absolute NaOH? Ans. 61.25 Gm. Bquation: 2 NaOH + H^SO* = Na2S04+2H20. 20. How much Na^SO^ is formed ? Ans. 88.75 Gm. 21. How much water is formed as a by-product? 22. How much Ca (PHjOj), is required to make 2 oz. of KPH2O2? Ans. 715.2 gr. Equation : Ca (PH^Oj) j-f K2COs=:2KPH202-f CaCO, 2^. How much CaCOg is formed as a by-product ? 24. How many Gm. of silver nitrate can be made from 100 Gm. of silver?* Ans. 157.477 Gm. Equation : 3 Ag -f 4HNO3 = 3 AgNOg+NO-j-a HjO. 25. How much NO will be evolved? Ans. 9.2 Gm. 26. How much absolute HNO3 will be used for the purpose? Ans. 77.85 Gm. 27. How much dried alum can be made from 50 Gm. of crystallized alum? Note. — The dried alum is alum minus the water of crystalliza- tion. 28. Hydriodic acid may be made by the interaction of potassium iodide and tartaric acid, as follows: KI -f H^C.H.Oe = HI 4- KHC,H,Oe. How much potassium iodide, and how much of tartaric acid, is required to make 50 Gm. of the hydriodic acid ? • In balancing oxidation and reduction equations employ Johnson's rule. 123 Pharmaceutical Arithmetic 29. One pound of baking powder is to contain 9 oz. of cream of tartar, and enough baking soda for neutraliza- tion.* How much of the latter is required? Ans. 4.02 oz. Equation : KHC^H.Oe+NaHCO, = KNaC^H,Oe+COa+HA 30. How much carbon dioxide would a pound of this baking powder yield? Ans. 2.1 oz. 31. A druggist has use for 200 gr. of lithium citrate. He has none in stock, but has the carbonate. So he decides to make the citrate from the carbonate by saturation with citric acid. How much of the carbonate should he use? Ans. 105.7 gr. Equation : 3Li3CO,+2H3C,H,0^2L:3C,HA4-3CO,+3HA 32. How much absolute mercuric oleate in 100 Gm. of official oleatum hydrargyri? How much uncombined oleic add? Note. — ^All HgO used is converted into oleate. Equation: HgO + sHQsHsaO, = Hg (C^sHj.O,), + H,0. 33. A druggist desires to make quinine bisulphate from the normal sulphate. How much absolute H2SO4 would be required by 100 Gm. of the normal salt? Ans. 11.256 Gm. Equation : (Cj oH, J^^02)^:,S0^.7'R/D + HjSO* +; 7H2O = 2CaoH,*N,0,HjS0^.7H,0. Cai.cui*ations Based on Reactions, but Involving A1.S0 Percentage Solutions. From the equation, 3Hg + 8HNO3 = 3Hg(NO,)j -f 2NO + 4H2O, the amount of absolute HNO, required to dissolve a given amount of Hg can be calculated by pro- portion. However, the acid which is actually used in the operation is not absolute HNO3, but is an aqueous solution *The rest is corn starch. Calculations Based on Reactions, 123 of the latter, 68% strong. A second calculation, in which it is determined how much 68% acid would be equivalent to [that is, would contain] the amount of absolute HNOg found by proportion, is therefore necessary, if the answer is to form the basis for practical work. 34. How many Gm. of nitric acid, 68% strong, are required to dissolve 20 Gm. of mercury as mercuric nitrate ? 3Hg + SHNOa = 3Hg(N03)a + 2NO ,+. 4HA given sought 601.8 504.16 601.8 parts : 504.16 parts :: 20 Gm. : x ( 16.75 Gm.). The answer indicates the amount of absolute HNO3; and from this answer the amount of 68% acid containing 16.75 Gm. of absolute HNOg may be found by a second proportion. See chapter VI. For this second proportion the three known terms are : 68%, 100% [the % of the abs. HNO3], and 16.75 Gm. The answer is to express weight. Hence — : :: 16.75 Gm. : x The answer is to be larger than 16.75 Gm., because the question is, how much of a weaker acid is equivalent to 16.75 Gm. of absolute HNO3. Hence — 68% : 100% :: 16.75 Gm. : x (24.63 Gm.). Accordingly, 24.63 Gm. of official nitric acid is required to dissolve 20 Gm. of mercury. [In practice an excess is used.] Note. — As has been stated before, dividing the divisor is equivalent to multiplying the dividend. It follows then that (16.75 Gm. X 100) -r- 68 = 16.75 Gm. -=- (68 -f- 100). In other words, the amount of weaker acid may be found by dividing the amount of absolute acid by the percentage strength of the weaker acid, expressed in hundredths. Thus — 16.75 Gm.-i- .68 =24.63 Gm. See page 70. 35. How many lb. of potassium acetate can be made from 50 lb. of acetic acid, 36% strong? Remarks. — This problem may be solved in two ways: cither the absolute acid in 50 lb. of the 36% acid may be 124 Pharmaceutical Arithmetic calculated, and this amount of absolute acid placed as the third term, in which case the fourth term will be the answer desired; or, the 50 lb. [of 36% acid] may be used as the third term, giving as the fourth term the amount of a 36^ solution of potassium acetate, from which, by a subsequent calculation, the amount of [absolute] potassium acetate may be found. Solution, the first-mentioned process being used. — 100% : 36% :: 50 lb. : x (18 lb.), showing that the 50 lb. of 36% acid contain 18 lb. of abso- lute acid. The problem has thus been simplified to — How much KCaHaOj can be made from 18 lb. of HCaHaOj? Equation : HC2H3O2 + KHCO, = KC2H3O, + CO2 -f H,0. given sought ^ 60.03 98.12 Then — 60.03 parts : 98.12 parts :: 18 lb. : x x=29.4 lb. = wt. of KC2H3O2. 36. How many Gm. of NajCOa.ioHjO must be used to make 500 Gm. of 5% solution of NaOH ? Solution. — 100% : 5% :: 500 Gm. : x (25 Gm.), showing that the 500 Gm. of solution contain 25 Gm. of NaOH. Equation: Na2CO,.ioHjO-l-Ca(OH)2=2NaOH + CaCO, + ioH20 sought given 286.16 80.02 80.02 parts : 286.16 parts :: 25 Gm. : x(89.4Gm.). ^y. A druggist desires to prepare 250 Gm, of dil. HBr (10%) by the tartaric acid method. How much KBr and how much H2C4H40e are required? Ans. 36.73 Gm. of KBr and 46.33 Gm. of the acid. Equation: KBr + H^C^H^Oe = HBr -j- KHC^H^O,. Calculations Based on Reactions 125 38. HCN may be made extemporaneously from AgCN and HCl. How much hydrochloric acid, 31.9% strong, would be required by i oz. of silver cyanide? Equation: AgCN + HCl = AgCl + HCN. 39. How many gr. of dil. HCN (2% strong) would be produced? 40. How much stearic acid, and how much crystal- lized sodium carbonate are required to make 500 Gm. of glycerin suppository mass, the latter to contain 4% of sodium stearate as stiffening agent? Solution'. 4% of 500 = 20 Gm. (amt. of sodium stear- ate required). Equation : sHCigHsoO^+Na^COa.ioHaO = 2NaCi8H3502+C02+iiH20 sought given 568.58 612.58 Then— 612.58 : 568.58 :: 20 Gm. : x R. wt. sod. stear. : R. wt. st. ac. : : wt. sod. st. : wt. st. ac. And in like manner the amt. of sodium carbonate is found. React, wt. sodium stearate : react, wt. sodium carbonate :: 20 Gm. : x. 41. How much HgClg is required to yield enough yellow mercuric oxide for j5/2 lb. of stock ointment 40% strong ?* Equation: HgCl^ + 2NaOH = HgO + 2NaCl + H^O. • If the right amount of the oxide Is precipitated, it is but necessary to wash, drain, and mix with enough wool fat to make % lb. of stock oint- ment, which may be quickly diluted as needed, yielding an ointment far superior to that made from the dried oxide by trituration. 126 Pharmaceutical Arithmetic PrOBUSMS iNVOtVINC ALSO REDUCTION 0:P VOLUMB TO Weight and Vice Versa. 42. How many grains of Zn would be converted into ZnClj by 200 c.c of hydrochloric acid, sp. gr., 1.16, strength, 30%? Solution.^ 200 cc [X 1.16] = 232 Gm. 100% : 30% : : 232 Gm. : x (69.6Gm.ofabs. HCl). Equation:— Zn + 2HCI = ZnCl, ±^ H,. sought given 65-37 72.94 Then— 72.94 parts : 65.37 parts :: 69.6 Gm. : x (62.4 Gm. of Zn). 43. How many gr. of LijCO, would be dissolved as LiCl by I f5 of hydrochloric acid, sp. gr., 1. 158, strength, 31.9% ? Ans. 171 gr. Equation', LijCOj + 2HCI = 2LiCI + CO, + HA 44. How many c.c. of nitric acid, sp. gr., 1.403, strength, 68%, can be made from 600 Gm. of NaNOj? Ans. 465.8 c.c. Hgwa^ion :—NaNOa + H^SO^ = NaHSO^ '+ HNO,. 45. How many c.c. of sulphuric acid, sp. gr., 1.826, strength, 92.5%, would be required by the 600 Gm. of NaNOa? 46. How many fj of ammonia water, sp. gr., .897 strength, 28%, could be neutralized with 500 cc. of acetic acid, sp. gr., 1.045, strength, 36% ? Solution. — Step I: Finding amount of absolute acetic acid. 500 X 1.045 (sp. gr.) = 522.5 Gm. (wt. of acid). Then 100% : 36% : : 522.5 Gm, : x (188.1 Gm. of ab- solute acetic acid.) Step II: Finding amount of abs. ammonia (NH,) this amount of abs. acid would neutralize. Calculations Based on Gravimetric Analyses 127 Equation:— NHj + HC2H3O2 = NH.QHsOa. sought given 17.03 60.03 60.03 • 17-03 " 188.1 Gm. : X x= 53.32 Gm. (am't of NH, neut. by 188. i Gm. of abs. acid). Step III : Finding Gm. of 28% ammonia water equal to 53.32 Gm. of abs. NHj. 28% : 100% :: 53.32 Gm. : x (190.43 Gm.). Step IV : Reducing weight in Gm. to volume in f 5. 190.43 -f- .897 (sp.gr.) =212.3 c.c. And 212.3 c.c. -7- 29.57 = 7.2 ^5- (Ans.) Problems Based on Gravimetric Analyses. Problems based on equations occur not only in manu- facturing, but also in analytical chemistry. For instance : 47. 1.2 Gm. of crude barium hydroxide is dissolved by aid of HCl ; the resulting BaClg precipitated as a sulphate — as BaSOi, ^^^ the latter washed, dried and weighed. Its weight is found to be 1.6 Gm. What % of Ba(OH)2 in the sample ? Equation: Ba(OH)2 + 2HCl = Baa2 + 2H2O BaCla + HjSO, = BaSO, + 2HCI. Solution: It will be seen that a molecule of BaS04 re- quires for its production a molecule of Ba(OH)2. Hence 233.44 parts (mol. wt. of BaSOi) oi the precipitate indicate 171.37 parts (mol. wt. of Ba(OH)2). And since all the Ba(OH)2 in the sample has been converted into BaSOi, the calculation is : 233.44 parts : 171.37 parts :: 1.6 Gm. : wt. of Ba(OH)2 in sample. Then weight of sample (1.2 Gm.) : wt. of Ba(OH)j in sample :: 100% : x. N. B.— Use the general rules for stating proportions. 128 Pharmaceutical ^^rithmetic 48. .6 Gm. of a certain iron ore is dissolved in HCl, oxidized to ferric iron with HNO3, precipitated as ferric hydroxide, ignited to ferric oxide, yielding of the latter .52 Gm. What % of iron in the ore ? Solution: All the iron in the sample (.6 Gm.) is found in the ignited precipitate, existing in the latter as FejOg. If we find the amount of iron (Fe) in the .52 Gm. of FcjOa, we have therefore also the amount in the .6 Gm. of ore. Mol.wt.FcjOa : moJ.wt.Fpa : : .52Gm. :x(Feinprecip.). .6 Gm. : X : : 100% : y. (% of Fe in ore). Ans. 60.5%. 49. .5 Gm. of a certain phosphate yield .13 Gm. of Mga P2O7. Calculate the % of phosphorus in the phosphate. 50. Calculate the % of PO^. 51. This would equal how much P2O5 ? * Note.— MgJP,0, = P, = 2 PO. = P^Og. Calculating the Analytical Factor, In the preceding problems the calculations were based upon combining weights, which formed the first couplet of the proportion in each case. If a large number of determinations of the same kind are to be made, the above calculations may be shortened by the use of factors. The analytical factor is simply the ratio between the first and second terms of the proportion based upon the combining weights, and is used as a multiplier of the final weight in the gravimetric analysis, giving as product the corresponding weight of substance to which the laboratory results are to be calculated. For instance : A number of phosphorus determinations are to be made, and the phosphorus is to be weighed as Volumetric Calculations 129 MgaPsOy. In the proportion the first couplet would be P, 62.08 Mg.P^O, : P, Mg^P^O, 222.72* If the ratio be found, 62.08 -i- 222.72 = .277, this may be used as a factor, and will give the amount of Pj in any known amount of MgaPaO^ by multiplication. If the weight of MgaPaO^ is .2 Gm., it will contain .2Gm. X .277 =.0554 Gm. of phosphorus. 53. Iron, Fcj, is to be determined as FejOg. What factor may be used ? , 54. CaO is to be determined as CaCOg. Calculate factor. 55. Use factor in calculating amount of CaO indicated by .2 Gm. of CaCOj. Problems P>ased on Volumetric Anai^yses. Normal Solutions. A normal, y, volumetric solution contains in each Liter the chemical equivalent of i Gm. of hydrogen. In other words, the quantity of the substance required to make i L. of normal solution is the molecular weight, expressed in Gm., and divided by the number indicating how many times the mole- cule is equivalent to H^. In case of acids, the basicity in- dicates the equivalence; and accordingly the molecular weight of HCl is divided by i (that is, is not divided at all) ; H2SO4 is divided by 2 ; H3PO4 by 3 ; etc. In case of bases, the valence of the metal or basylous radical gives the divisor. Thus, NaOH is divided by i, Ba(OH)2 by 2, etc. In case of precipitants, the valencies of the radical which recurs in the precipitate shows the equivalence. Thus NaCl, when it precipitates AgCl, equals CI', hence NaCl = Hj. The fol- lowing substances are equivalent to Hj : — HCl, NaOH, KOH, I, Br, KSCN, AgNOg, NaCl, Na^S^O,. X30 Pharnuiceutical Arithmetic The following are equiv. to H, : HjjS04, HaCjO*. 2HjO, Ba(OH)2, BaClj. KjCr^Oj = O3 ; hence = H^. 2KMn04 = O5 ; hence = Hio. Problem. — How much absolute H^SO* is required for iL. of ^H^SO,? Solution.— Mol. wt. H^0« = 98.09 .*. 98.09 Gm. -5- 2 = 49.045 Gm. (Ans.). Adjusting Strength of Normal Solutions. All normal solutions will, if they react, be equal to each other volume for volume. If then, one y solution of stand- ard strength is at hand, a normal solution of a reagent \^p- able of reacting with the solution on hand, can be made by comparison, that is, can be standardized against the first solution. For instance: A certain solution of HjSO^ requires for the neutralization of 10 c.c. 12 c.c. of y KOH. To what volume must 400 c.c. of the acid solution be diluted to make I Iv. of f strength ? Solution. — ^If 10 vol. of the H2SO4 solution required 12 vol. of ? KOH, then the 10 vol. contain as much HjSO, as should be contained in 12 vol. of ^ H:,S04, and 10 volumes must be diluted to 12 volumes. Hence— 10 vol. : 12 vol. :: 400 c.c. : x (480 c.c), which is the volume to which the 400 c.c. should bo diluted. See chap- ter vll. Fractional Normal Solutions. There are also solutions ^, -J, -^^ -^ and xhr ^^ normal strength, and are known as half-normal, fifth-normal, deci- or tenth-normal, fiftieth-normal, centi- or hundredth-normal respectively. Volumetric Calculations I31 Volumetric Factors. The normal factor of a substance is the amount in i c.c. of a Y solution of that substance. Thus the normal factor of H2S04 = (98.09 Gm.-7-2)-^ 1 000 = .049045 Gm. The deci-normd factor is the amount in i c.c. of a deci-normal solution. In like manner we may define f , ?. f^, and 1^ factors. As all Y solutions are equivalent c.c. for c.c. (if they react), it follows that the normal factor of a substance is al- so the amount indicated by i c.c. of a y solution of the opposite kind. For instance: i c.c. f HCl will just neutralize i c.c. of X NaOH. Then if a quantity of NaOH is just neutralized by I c.c. of Y HCl, it must contain as much absolute NaOH as I c.c. of y NaOH solution. That is, the y factor in- dicates the amount of abs. NaOH in the sample. Upon this deduction is based the method of calculating results from volumetric analyses. All this applies, of course, also to fractional normal solutions and to fractional normal factors. Problems. 56. Calculate the amount of AgNOg in i L. of f AgNOg solution. 57. Calculate the amount of K2Cr207(= He) for 500 c.c. of solution. 58. Calculate the amount of H2SO4 in 200 c.c. of | solution. 59. Calculate the f factor for H2C2O4.2H2O. Solution. — Mol. wt. = 126.05. A dibasic acid, hence = Ha .'. (126.05 H- 2) H- 1000 = .063025, n factor. 60. Calculate f factor for (a.) KOH, (b.) AgNO,, (c.) Ba"Cl2, (d.) K2Cr20,. 61. Calculate the ^ factor for (a.) I, (b.) NaoSjOs, (c.) 2KMnO, (=:H,o). 62. Calculate the -^ factor for strychnine, C21H22N2O2, equivalent to H^. Ans. 0.00668. 132 Pharmaceutical Arithmetic Use of Normal and Fractional Normal Factors. Suppose I Gm. of a commercial sulphuric acid requires for neutralization 20 c.c. of ^ KOH. Then the i Gm. of acid contains as much absolute H2SO4 as 20 c.c. y solution ; that is, y factor of 112804X20= .049045 Gm. X 20=. 9809 Gm. It will be observed that the factor to be used is the factor of the substance to be determined, — not the factor of the volumetric solution used in the titration. Problem. — 2.5 Gm. of commercial ferrous sulphate is dissolved to make 60 c.c. of solution. 12 c.c, which is 1-5 of this solution, require 17 c.c. of ^ potassium perman- ganate for complete oxidation, (a.) Calculate amount of FeS04.7H20 in sample, (b.) Calculate to percentage. Solution. — Mol. wt. of FeS0^.7H,0 = 278. PeSO^.THjO = H, .'.278-^-10,000 = .0278 Gm. (ft factor). And .0278 Gm. X 17 = .4726 Gm. (amt. of FeS0i.7H:,0 in 1-5 of sample weighed out, i. e., in .5 Gm.) Then .5 Gm. (wt. of test sample) : .4726 Gm. :: 100% : (x) 94 5%. CALCULATING A CORRECTION FACTOR. Some volumetric solutions, like deci-normal KMnO^, become weaker on keeping. Instead of fortifying to standard strength, which is very difficult, a correction factor may be calculated, and used as a multiplier of the volume of volumetric solution used. Suppose a KNInO^ solution has weakened so that 10 cc oxidize only as much iron as 9 c.c. of deci-normal KMnO^ so- lution. If then the volume of volumetric solution used in a titration is first multiplied by .9, the deci-normal factor of the substance tested may be employed in the usual way. The cor- rection factor is therefore a number which expresses the strength of a solution, with reference to the normal or fractional normal strength, which is taken as unity. Example. — A solution of Ba(OH)« originally of deci-normal strength, has weakened owing to absorption of COj, so that 10 c.c. of it neutralize only 9.2 c.c. of deci-normal HCl. This Ba(OH), solution is used in the titration of commercial acetic acid .5 Gm of which required 20 c.c. of the Ba(OH)a solution. Calculate amount of absolute HCjHjOj in the acid. Solution. — Since 10 c.c. of Ba(0H)2 solution = 9.2 of deci- normal solution, 20 c.c. of Ba(OH)j=18.4 c.c. of deci-normal Ba(OH).."..006 Gm. (deci-normal factor HC,:^0,) X 18.4 = .1104 Gm. (amt. o± abs. acetic acid). CHAPTER X. REDUCING VOLUMES OF GASES TO WEIGHTS AND VICE VERSA. If the molecular weight of a gas, based upon 0=i6, is expressed in Gm., and this weight of the gas is measured, under standard conditions of temperature and pressure, the volume is in every case about 22.4 L. Thus, 32 Gm, of O2, 2 Gm. of H2, 28 Gm. of Ng, each measure 22.4 L. ; so do 48 Gm. of O3 (ozone), 44 Gm. of CO2, and 17 Gm. of NH3. This is in conformity with Avogadro's law, that all gases contain the same number of molecules in equal volumes, under identical conditions of temperature and pressure. Accordingly, the weight of a given volume of any gas (of known molecular weight) may be calculated; so, also, the volume of a given weight. This may be done by proportion. As the weights and volumes are directly proportional (4 Gm. of H measuring twice as much as 2 Gm.) the general proportion for the determination of weight from volume would be: 22.4 L. : observed vol. in L. : : mol. wt. of gas in Gm. : x. x = wt. in Gm. of observed volume. And the general proportion for the calculation of the volume of a definite weight of any gas would be : mol. wt. of gas in Gm. : observed vt't. in Gm. : : 22.4 L. : x x = vol. of gas in L. The general rule for the statement of the proportion may be used, as in the following examples. I. What is the volume of 500 Gm. of oxygen? Solution. — Mol. wt. of 0 = 16X2 = 32. 134 Pharmaceutical Arithmetic Then the three known terms for the proportion are: 32 Gm., 500 Gm. and 22.4 L. The answer is to express volume. Hence — :: 22.4 L. : x The values being directly proportional, and the weight for which the volume is to be found being larger than the weight for which the volume is known [the molecular weight, expressed in Gm.], the answer will be larger than the third term. Hence — 32 Gm. : 500 Gm. :: 22.4 L. : x (350 L.)- 2. What is the weight of 50 L. of NH3 ? Solution. — Mol. weight of NH3 = I7. Then the three known terms are — 17 Gm., 50 L., and 22.4 L. The answer is to express weight. Hence — : :: 17 Gm. : x The answer is to be larger than the third term. Hence — 22.4 L, : 50 L. :: 17 Gm. : (38 Gm.). Problems. 3. What is the volume of 1 Kg. of O ? of 50 Gm. of CO2? of 30 Gm. of H2S? of 5 Gm. of CI? 4. What is the weight of 50 L. of SOj? of 10 L. of Nj.O (laughing gas)? of i gallon of O? of 300 c.c. of H? >fote. — Oxygen, chlorine, and hydrogen have two atoms in the molecule; hence the molecular weight is, for each of these elements, double the atomic weight. Efi'ect op Variation in Pressure on the Volume 0^ a Gas. When the pressure upon a gas is increased, the vol- ume of the gas is found to decrease correspondingly. II the pressure is doubled, the volume of the gas is thereby reduced to one-half the initial volume; if the pressure is tripled, the volume of the gas is reduced to one-third; if Effect of Pressure on Volume of a Gas 135 the pressure is reduced to one-half the initial pressure the volume of the gas becomes twice the initial volume; etc. Boyle's Law. — ^The relation of pressure to volume is stated by Boyle as follows: With the temperature con- stant, the volume of any given weight of any gas varies inversely as the pressure upon it. It follows then that if the volume has been observed at any given pressure, the volume which the gas would occupy at any other pressure may be calculated by pro- portion. 5. The NO gas, evolved from a certain amount of ethyl nitrite, measures 60 c.c. at a pressure of 744 mm.* How much would it measure at 760 mm ? Solution. — ^Volume is asked for. Hence — : : : 60 c.c. : x The values being inversely proportional, the NO would measure less at 760 mm. than at 744 mm. The an- swer is therefore to be smaller than the third term. Hence — 760 mm. : 744 mm. : : 60 c.c. : x (58.73 c.c.) Problems. 6. A Liter of ammonia gas at normal pressure, would measure how much at a pressure of two atmos- pheres? Ans. .5 L. 7. A quantity of CO2, which measures 84.5 c.c. at 751 mm., will measure how much at normal pressure? 8. A quantity of O, measuring 3 gallons at 760 mm., would measure how much at 1.520 M? Ans. 15^ gal. 9. What would be the volume of 3 Gm. of SOj at 748 mm. Note. — Remember that the volume of the "gramme-molecule" is 22.4 L. at normal pressure only. •Note.— Pressure of gases is expressed in units of length, the expression hav- ing reference to the height of a column of mercury which will have an equal pressure. Thus, a pressure of 760 mm. is a pressure capable of supporting (hence equal to) a column of mercury 760 mm. high. 760 mm. is the atmos- pheric pressure under standard conditions; and is known as normal pressure,. or aa one atmosphere. Two atmospheres would therefore equal 760 mm. X 2, 136 Pharmaceutical Arithmetic 10. What is the weight of 500 c.c. of CI, measured at a barometric pressure of 752 mm. Ans. 1.554 Gm. Note. — ^When a gas is collected over a liquid, as N,0 ovei water, the pressure of the gas is equal to the atmospheric pres- sure— ^i. e., barometric pressure. Ej'F'ect of Variation in Temperature on the Volume OF A Gas. The standard temperature for measuring gases is O^C. If the volume is taken at a different temperature [for sake of convenience], the volume at o°C. must be calculated from the observed volume before the constant, 22.4 L., can be made use of. With the pressure remaining unchanged, the volume of a gas varies directly as its absolute temperature.* By absolute temperature is meant temperature reck- oned from 273°C. below o°C. In other words, the zero of absolute temperature is minus 273°, or — 273°C. It follows then that temperature in °C. may be reduced to absolute temperature by adding 273°!- Thus, I5°C. is 273 -|- 15 = 288°C. absolute temperature; and ioo°C. is 273 -f- 100 = 373 °C absolute temperature. 11. A pharmacist in assaying spt. of nitrous ether has obtained a volume of NO gas measuring 56 c.c. at a temperature of 22 °C. What is the volume of the gas at o°C.?t Solution. — 22°C = 273 -\- 22 = 295 °C. abs. temp. o^C. = 273 -{- o = 273 °C. abs. temp. The three known terms are: 295*'C [absolute tem- perature at which volume was observed], 273 °C [abso- lute temperature at which volume is wanted], and 56 c.c. [observed volume.] The answer is to express volume. Hence — : : : 56 c.c : x As the volume varies directly with the temperature; •This "law" is not exactly true for all gases. tFor directions for adding a positive quantity to a negative quantity (tor instance, 273°C. to — 20°C.), see chapter XI. Jin the U. S. P., Qth Bev., the gasometric assays have been simplifled by the use of factors based on volume of gas at 25°C. But the methods of gen- eral gasometric analysis involve the above calculations. Bffect of Temperature on Volume of a Gas 13^ and as the temperature at which the volume is wanted is lower than the temperature at which the volume is known, the answer must be smaller than the third term. Hence — 295°C : 273*0 : : 56 c.c. : x(5i.8 c.c.) The general proportion would be : Absolute temperature at which volume was observed is to absolute temperature at which volume is wanted [273 °C + o°C. = 273 "C] as observed volume is to x.* Problems. 12. A quantity of CO, measures 35 c.c. at 2o''C. What will it measure at o°C.? Ans. 32.6 c.c. 13. A quantity of hydrogen measures 5 L. at 23° C. What would be its volume at o°C.? What at t;o°C? What at 100° C? 14. A quantity of NO measures 60 c.c. at a pressure of 750 mm., and at a temperature of 20° C. What would be its volume at normal pressure and temperature. Solution. — 760 mm. : 750 mm. : : 60 c.c. : x X = 59.21 c.c. = vol. at normal pressure. Then 273 + 20°C. : 273 + o°C. : : 59.21 c.c. : x X := 55- 16 c.c. = vol. at normal pressure and temperature. 15. What is the weight in Gm. of 60 c.c. of NO gas, measured at a barometric pressure of 750 mm., and at a temperature of 20° C? Solution. — Having found the volume of the gas at normal pressure and temperature, as shown in the pre- ceding example, we calculate the weight of that volume [in this case, 55.16 c.c] of NO thus: 22.4 L. : .05516 L. :: 30 Gm. [mol. wt. of NO] : x. x =: .074 Gm. ' 16. How much C2H5NO2 [ethyl nitrite] is indicated by 60 c.c. of NO at a pressure of 750 mm., and at a tem- perature of 20° C? It will be seen that a gas at CO. expands ,^3 of its volume for an increase ffi temperature of 1°C. ; and that it contracts ^tf, of its volume for a decrease ia Uiperature of 1°0. ; in short, that the coefficient of expansion of any gas iSsHi or, decimally, .OttseeS. ijS Pharmaceutical Arithmetic Solution. — Having calculated the weight of the NO, we calculate the amount of CgHgNOa required to produce the weight [in this case, .074 Gm.] of NO gas.* Equation : QH.NO, + KI + H2S04 = NO + C3H60H + KHS04 + I. sought given 7505 30 Then — 30 parts : 75.05 parts :: .074 Gm. : x (.185 Gm.). 17. How many c.c. of CO2, at normal pressure, and at a temp, of 25 °C., would be generated from .5 Gm. of NaHCOg? Equation: 2NaHC03 + H^SO^ = Na2S04 + 2CO2 f- 2H,0. 18. How many L. of hydrogen [pressure, 748 mm., temp., 20° C] could be made from 5 Gm. of zinc? 19. How much KCIO3 is required to generate 25 L. ofO.? Equation : 2KCIO3 = 2KCI + 3O2. Note. — ^If pressure and temperature are not given, nomal pressure and temperature are understood. 20. How much NH4NO3 is required to make 10 gallons of N2O [laughing gas] ? Equation: NH^NOg = N2O + 2H2O. 21. How much NH4NO3 is required to make enough NgO to fill a 10 gallon tank, the temperature being 20°C., and the barometric pressure, 750 mm.? Fbr equation see problem No. 20. 22. A certain quantity of baking powder yielded 70 c.c. of CO2; — temp., 20 °C., pressure, 748 mm. How much NaHCOg did the quantity of baking powder con- tain? Ans. .24 Gm. Equation: NaHCOg -f acid = COj etc. 24. How many c.c. of ammonia water, sp. gr. .96^ strength, 10%, would contain 5L. of NHg gas? *Tu the n. S. P.. 9tb Bev.. a factor is used to simplify ibis oakuktion. Sue factors" i>a?e 128. CHAPTER XI. THERMOMETER SCALES. On a Centigrade thermometer the temperature ol melting ice [freezing point of water] is zero; and the temperature of steam, at normal pressure [boiling point of water], is loo. On a Fahrenheit thermometer the temperature of ^Z' 5? do ee ?o 60 SO 4C dO 20 » /a so /07.& \ eoo I60 .i40 /MO IM 60 60 m*tnf*9 ^o/frr OiXB Uo o to 40 i eo TO •o -40- fo to o 10 \ao 40 I 240 Fharmaceutical Arithmetic rjnelting ice is 32 above zero ; and the temperature of steam, at normal pressure, is 212. On a Reaumur thermometer the temperature of melting ice is zero ; and that of steam, at normal pressure, is 80. On all three thermometers the graduation is con- tinued above the boiling point, and below the zero point. Temperature in degrees below zero, on any scale, is ex- pressed as a negative quantity, the minus sign being placed before the number thus, — 45°, which means 45 '^ below zera If the minus sign is absent, the number is understood to be a positive number, and to denote degrees above zero. Centigrade to Fahrenheit. It will be seen that 100 ''C = i8o°F. = 8o**R.; and hence that i°C. = 1.8° F = .8°R. ProW^m.— Convert 6o°C. into °F. Solution.— Since i°C. = i.8°F., 6o°C = 60* X 1.8 = io8°F. But since 6o°C. means 60° above the freezing point, the io8°F. must be counted from that same point ; and as the freezing point is marked 32 on the F. scale, 32 must be added to 108 in order to get the reading count- ing from zero F. Accordingly, 108° -|- 32° = 140° ; and 6o°C. = I40''F. Whence the rule— (°C. X 1.8) + 32 = °F. Problem.— Comert — 25°C. into "F. Solution. — 25 X 1.8 = — 45. [Remember that when a negative quantity is multiplied by a positive quantity, the product is a negative quantity.] It will thus be seen that 25 divisions on the C. scale equal 45 on the F. scale. But since the freezing point on the F. scale is 32* above zero, and since the °C. are counted from the freezing point, the temperature of — ^25 "C. would be — ^45" + 32; which means that the reading would he 5^° higher on the F scale than — ^5", hence would be [ — ^45 + 32] — I3**F. The rule for adding a positive quantity to a negatl^' THermometer Scales I41 quantity is as follows : Subtract the smaller number from the larger, and read the remainder with the sign of the larger number. Thus, — 10 -|- (-{- 32) = + 22; —42 + (+ 32) = —10. Pro^/^m.— Convert — lo^C. into °F. Solution. —10° X 1.8 = —18°. And —18 + (+32) = +14. Hence — io°C = I4°F. Note.— To add 32 means that the reading is to be 32° higher on the scale. Fahrenheit to Centigrade. Pro&/^w.— Convert 6o°F. into °C. Solution. — The first step to be taken is to find what would be the reading on the Fahr. thermometer if the freezing point were zero. This reading would be 32° lower than 60°, for zero on the Fahr. thermometer is 32° below the freezing point. Accordingly, we subtract 32" from 60°. We have now ascertained that the temperature of 60** F. equals 28° assuming the zero point to be where it is on the Centigrade thermometer, namely at the freezing point. Now i°C. = 1.8° F. ; then 28° F. must equal as many degrees C. as 1.8 is contained in 28°. Accordingly, we di- vide 28° by 1.8: 28'' -f- 1.8 = 15.5° C. Whence the rule — ("F. — 32) -V- 1.8 = "C. Pro&/^w.— Convert 10° F. into "C. Solution. — If the zero on the Fahr. thermometer were at the freezing point [where it is on the Centigrade ther- mometer], the reading would be 32" lower than 10", hence would be — ^22°. Then —22* h- 1.8 = — I2.2*C. (Ans.) Problem.— Convert — lo^F. into °C. 142 Pharmaceutical Arithmetic Solution. — If the zero on the Fahr. thermometer were at the freezing point [as in the Centigrade scale], the read- ing would be 32" lower on the scale than — 10°, hence would be — ^42°. Then —42" ~ 1.8 = —23.3° C. Note. — It will be seen that (+32) is subtracted from a posi- tive quantity [from an above zero reading] smaller than 32 by finding the difference between that positive quantity and 32; that difference being a negative quantity, i. e., a reading beyond [below] the zero point. And that (+32) Is subtracted from a negative quantity [from a below zero reading] by adding 32 to the negative quan- tity, and calling the sum a negative quantity. In short, to subtract 32 means to find the reading 32° lower on the scale. Centigrade and Reaumur.* Profc/^w.— Convert I5°C. into **R. Solution.— 100" C. = 80° R. Then I'^C. = .8" R, And is^C. = 15 X .8 = 12'' R. Whence the rule — "C. X .8 = ''R. Problem.— Convert 15" R. into °C. ^o/M/«on.— Since .8°R.— i.°C.,i5°R.— i5*'-t-.8=i8.7*'C. Whence the rule — OR. ^ .8 = "C. Fahrenheit and Reaumur. Pro&/^w.— Convert 6o°R. into °F. Solution.— Since 8o°R. = 180° F., i°R. = 2.25° F. And 60° R. = 60 X 2.25 = I35°F. But since the zero of the R. scale is at the freezing point, the I35°F. must b€ counted from the same point. Hence 135° + 3^** = i67°F. Whence the rule — (°R. X 2.25) + 32 = "F. •The Reauraur scale is not used in the United States, and is of interest only because temperature is expressed in °B. in some European publications. Thermometer Scales 143 Problem.— Convert 61 °F. into "R. Solution.— 61 °F. = 61 — 32 = 29°F. above the freez- ing point. Then 29** -j- 2.25 = I2.2°R, Whence the rule — (°F.~32) -=- 2.25 = °R. Problems. 1. Convert (a.) 25°C. into °F.; (b.) 8o''C. into "F. 2. (a.) 5X into °F. ; (b.) — 5°C. into °F. 3. (a.) — 30°C. into °F.; (b.) 2.5°C. into °F. 4. (a.) — 2.5°C. into °F.; (b.) i.i°C. into "F. 5. (a.) I25°F. into °C.; (b.) 25°F. into °C. 6. (a.) S^F. into °C.; (b.) 5.8°F. into °C. 7. (a.) — 5°F. into °C. ; (b.) — i.i°F. into *C. 8. (a.) — 25°F. into "=0. ; (b.) 31° F. into °C. 9. (a.) io°C. into °R. (b.) io°R. into °C. 10. (a.) — io°C. into °R. (b.) 10'' C. into "R. 11. (a.) — io°R. into "C. (b.) i.i^R. into "C. 12. (a.) 25°F into °R. (b.) 25°R. into °F. 13- (a.)— 25°F. into °R. (b.)— 25°R. into "F Answers. I. (a.) 77°r.; (b.) i76^F. 2. (a.) 41"?.; (b.) 23»F. 3. (a.) -22"^; (b.) 36.5°F.4. (a.) 27.5°F.; (b.) 33.98"?. 5. (a.) 5I.66X. 6. (a.) — isX. 7. (a.) — 20.5X. 8. (a.) — 3i.6"'a 144 PharmaceuticcU Arithmetic MISCELLANEOUS. BEAUME HYDROMETER SCALES. (a.) For Liquids heavier than water. Rule: Subtract the observed degrees Beaum^ from 14S, and divide the remainder into 145. The answer is the sp. gr. Sp. gr. = 145 -^ (145 — "B) Prc^blem. — ^A certain carboy of hydrochloric acid is marked 20° B, What is its sp. gr.? Calculation.— 145 — 20 = 125; 145 -^ 126 = 1.16 (sp. gr.) (b.) For liquids lighter than water. Rule: Add the observed degrees Beaum^ to 130, and divide the product into 140. The answer is the sp. gr. Sp. gr. = 140 -^ (130 + °B) Problem- — Ammonia water, marked 26° B., has what spw Calculation.— 130 + 26 = 156; 140 -^ 156 = .8974 sp. gr. PR0BUEM8. 1. Reduce to sp. gr. the following degrees B. on scale for liquids heavier than water:— (a.) 18°B.; (b.) 25°B.; (c.) 30°B.; (d.) 50°B. 2. Reduce to sp. gr. the following degrees B. on scale for liquids Ughter than water;— (a.) 18°B.; (b.) 30°B.; (c.) 40°B.j (d.) eO°B. ANSWERS. L (a.) 1.00; (b.) 1.2; (c.) 1.28. 2. (a.) .946; (b.) .874. CALCULATING DOSES FOR CHTLDKEN. The rule most generally employed is known as Dr. Young's Rule, and is as follows: Multiply tne adult dose by the fraction obtained by dividing the age of the child [in years] by its age plus 12. Problem. — If the adult dose of the medicine is 5 gr., what is the dose for a child 4 years old? Calculation— 4 -r- (4 + 12) = .^ = %. Then 5 gr. X y4 = I = IV* gr. Problem. — The adult dose of a certain medicine is % gr. How much should be given to a child 2 years old? Calculation.— 2 -r- (2 + 12) = 3^ = J Then i/s gr. X ^=5Vgr. Note. — Children are very susceptible to narcotics, especially to opium; and of these medicines less than one-half the amoimt in- dicated by Young's Rule is usually given. Of cathartics, howeveiv double the calculated d»se is generally required. Reference Tables 145 REFERENCE TABLES. VOLUME AND WEIGHT.* A. Volume to Weight. 1. Cubic Centimeters to Grammes. no. of c.c. given X sp. gr. = wt. in Gm. 2. Fluid Ounces to Grains. (454.6 X sp. gr.) X no. of f 5 given = wt, in gr. 3. Minims to Grains. (.95 gr. X sp. gr.) X no. of Til given = wt. in gr. 4. Cubic Centimeters to Grains. (no. of c.c. given X sp. gr.) X 15.432 = ■wt. in gr. 5. Fluid Ounces to Grammes. (no. of fS given X 29.57) X sp. gr. = no. of Gm. B. Weight to Volume. 1. Grammes to Cubic Centimeters. (no. of Gm. given -r- sp. gr. = vol. in cus. 2. Grains to Fluid Ounces. no. of gr. given -r- (454.6 X sp. gr.) = vol. in fj. 3. Grains to Minims. no. of gr. given -r- (.95 gr. X sp. gr.) = vol. in Ht. 4. Grammes to Fluid Ounces (no. of Gm. given -^ sp. gr.) -r- 29.57 = vol. in fj. 5. Grains to Cubic Centimeters (no. of gr. given X .0648) -r- sp, gr. = no. of c.C VALUES WHICH ARE PROPORTIONAL. Quantity (in wt. or vol.) to money value. In weight-percentage. % to weights, but not to volumes. In volume-percentage. % to volumes, but not to weights, in >^ solutions, (a.) gr. of constituent to Til of solution. (b.) Gm. of constituent to c.c of solution. In concentration and dilution . % inversely to wt. of solution. In alligation problems. Proportionate parts to weights. In chemical problems, atomic and molecular weights to actual weights. In gasometric problems. (a.) volumes of gases to their weights. (b.) volume inversely to pressure. (c.) volume directly to absolute temperature. In figuring interests, profit and loss, discount, etc., % is directly proportional to money value. * Remember that a cubic centimeter is a mil. Table of Atomic Weights Element Alnminmn. Antimony.. Arsenic Barium Bismuth.... Boron Bromin Cadmium... Calcium.... Carbon Chlorin Chromium. Cobalt Copper Fluorin Gol(i Hydrogen . lodin Iron Al Sb As Ba Bi B Br Cd Ca C CI Cr Co Cu F Au H I Fe At. Wt. Based on 0 = 16 27.1 120.2 74.% 137.37 208.0 11.0 79.92 112.4 40.07 12.0 35.46 52.0 58.97 63.57 19.0 197.2 1.008 126.92 55.82 Approx imate At. Wt 27 120 75 137 208 11 80 112 40 12 35 52 59 64 19 197 1 127 56 Element Lead Lithium Magnesium .. Manganese... Mercury Molybdentmi Nickel .... Nitrogen Oxygen Phosphorus.. Platinum Potassium.... Silicon Silver Sodium Strontium.... Sulphur Tin Zinc Pb Li Mg Mn Hg Mo Ni N O P Pt K Si Ag Na Sr S Sn Zn At. Wt. Based on 0 = 16 207.1 6.94 24.32 54.93 200.6 96.0 58.68 14.01 16.0 31.04 195.2 39.1 28.3 107.88 23.0 87.63 32.07 119.0 65.37 Approx imate At.Wt. 207 7 24 55 200 96 59 14 16 31 195 39 28 108 23 88 32 119 65 146 APPENDIX Nearly sixty-five per cent, of all the problems in the book are " answered " in the text. It has been thought best to supply in this appendix an additional list of answers, some " approximate." CHAPTER I No. 21 : 9.36 Gm., 5.76 Gm., 7.2 Gm. ; No. 55: 6 fi. 5; No. 74: 145+. CHAPTER II No. 13: i inch smaller; No. 49 (b) : 5.41 mg.; No. 54: 56.7 Gm., 56.7 Gm., 28.35 Gm., 425.25 Gm. CHAPTER III No. 10: Sp.gr,, .908; No. 40: 175.7 C.C.; No. 42: 258.6 c.c; No. 47: 101.25 c.c; No. 51: 1. 14 fl. 5; No. 53: 4.4 fl. 5; No. 54:343pts.; No. 63:21.5^. CHAPTER V No. 4 (b) im, (c) 4.67^, (d) 4.3<^; No. 7: $5.25; No. 8 (a) : 63.4^, (b) 14^, (c) 71^, (d) 41.6 gal.; No. 16: $21.84. CHAPTER VI No. 6: 18%; No. II: 8,000 Gm. ; No. 13: 1.2%; No. 15: 36.36%; No. 17: 5.88% ; No. 24: .96 oz. and 31.04 oz.; No. 25: 3996 Gm. ; No. 31:1 13.5 Gm. ; No. 32 : 292.2 Gm. ; No. 37 (b) : 3378 Gm. ; No. 38: 42.5% ; No. 40 : 26.85% J No. 53 : 1.05 Gm. of apomorphine; No. 54: 3.83 Gm. cocaine. CHAPTER VII No. 8: 2219.5 Gm., No. 9: 2399.71 Gm.; No. 10: 3.45 lb.; No. 14: 716.25 Gm.; No. 21 : 4.44 Gm.; No. 22: 2.5 Gra. CHAPTER VIII No. 14 : 84 parts abs., 40 parts 36%, 40 parts water ; No. 19 : 125 Gm.; No. 20: looo c.c; No. 27: 15.3+ gr- and 484.6+ gr- CHAPTER IX No. 7: 59.85%; No. 9: 75%; No. 12: 69.8%; No. 14: 45-3%; No. 18: 43+ Gm.; No. 21 : 22.5 Gm.; No. 28: 64.9 Gm. of KI and 59.2 Gm. of tartaric acid ; No. 41 : 4 oz ; No. 45 : 409 c.c. CHAPTER X No. 3 : 699.78 L. of oxygen, 25.4 L. of CO,; No, 4: 143.5 Gm. of SOj. 147 8EE NEXT FAa7 BY THE SAME AUTHOR A concise, systematic course of study- in the branch of technical Latin of interest to students of pharmacy and of medicine. A course in pharmaceutical Latin nomenclature and prescription writing and reading. But so planned that it may be taken coincident with the junior courses in materia medica and pharmacy. Based on the Pharmaceutical Syllabus Requirements. SECOND EDITION, . . $1^0 Published by the Author J. W. STURMER Philadelphia College of Pharmacy o ^1 oil -p; a X c 0 i. University of Toronto Library DO NOT REMOVE THE CARD FROM THIS POCKET Acme Library Card Pocket Under Pat. "Ref. Index File" Made by LIBRARY BUREAU